Differentiation

If $x=a \sin pt$, $y=b \cos pt$ find $\dfrac{d^2 y}{dx^2}$ at $t=0$

$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ $\cdots$ (1)

$\dfrac{dy}{dt}=-bp\sin \left(pt\right)$ $\cdots$ (2)

$\dfrac{dx}{dt}=ap \cos \left(pt\right)$ $\cdots$ (3)

$\therefore$ In view of equations (2) and (3) equation (1) becomes

$\dfrac{dy}{dx}=\dfrac{-bp \sin \left(pt\right)}{ap \cos \left(pt\right)}$

i.e. $\dfrac{dy}{dx}= \dfrac{-b}{a} \tan \left(pt\right)$

$\therefore$ $\dfrac{d^2 y}{dx^2}=\dfrac{d}{dx}\left[\dfrac{-b}{a} \tan \left(pt\right)\right]=\dfrac{-b}{a}\times p \times \sec^2 \left(pt\right) \times \dfrac{dt}{dx}$ $\cdots$ (4)

From equation (3), $\dfrac{dt}{dx}= \dfrac{1}{dx/dt}=\dfrac{1}{ap}\sec \left(pt\right)$ $\cdots$ (5)

Substituting the value of $\dfrac{dt}{dx}$ from equation (5) in equation (4) gives

$\dfrac{d^2 y}{dx^2} = \dfrac{-b}{a} \times p \times \sec^2 \left(pt\right)\times \dfrac{1}{ap}\times \sec \left(pt\right)$

i.e. $\dfrac{d^2 y}{dx^2} = \dfrac{-b}{a^2}\sec^3 \left(pt\right)$

$\therefore$ $\dfrac{d^2 y}{dx^2} \bigg|_{t=0} = \dfrac{-b}{a^2} \sec^3 \left(0\right) = \dfrac{-b}{a^2}$

Differentiation

Find $\dfrac{dy}{dx}$ if $y = \left(x\right)^{\cos x} + \left(\sin x\right)^{\tan x}$

Let $y = p\left(x\right) + q\left(x\right)$ where $p\left(x\right) = \left(x\right)^{\cos x}$ and $q\left(x\right)=\left(\sin x\right)^{\tan x}$

Then $\dfrac{dy}{dx} = \dfrac{dp}{dx} + \dfrac{dq}{dx}$

$p = \left(x\right)^{\cos x}$

$\therefore$ $\log p = \log \left(x\right)^{\cos x} = \cos x \log \left(x\right)$

$\therefore$ $\dfrac{1}{p} \dfrac{dp}{dx} = \dfrac{\cos x}{x} - \sin x \log \left(x\right)$

$\therefore$ $\dfrac{dp}{dx} = p \left[\dfrac{\cos x}{x} - \sin x \log \left(x\right)\right]$

i.e. $\dfrac{dp}{dx} = \left(x\right)^{\cos x} \left[\dfrac{\cos x}{x} - \sin x \log \left(x\right)\right]$

$q = \left(\sin x\right)^{\tan x}$

$\therefore$ $\log q = \log \left(\sin x\right)^{\tan x} = \tan x \log \left(\sin x\right)$

$\therefore$ $\dfrac{1}{q} \dfrac{dq}{dx} = \dfrac{\tan x}{\sin x} \times \cos x + \sec^2 x \log \left(\sin x\right)$

i.e. $\dfrac{1}{q} \dfrac{dq}{dx} = 1 + \sec^2 x \log \left(\sin x\right)$

$\therefore$ $\dfrac{dq}{dx} = q \left[1 + \sec^2 x \log \left(\sin x\right)\right]$

i.e. $\dfrac{dq}{dx} = \left(\sin x\right)^{\tan x} \left[1 + \sec^2 x \log \left(\sin x\right)\right]$

$\therefore$ $\dfrac{dy}{dx} = \left(x\right)^{\cos x} \left[\dfrac{\cos x}{x} - \sin x \log \left(x\right)\right] + \left(\sin x\right)^{\tan x} \left[1 + \sec^2 x \log \left(\sin x\right)\right]$

Differentiation

Find $\dfrac{dy}{dx}$ when $y=x^{\cot x} + \dfrac{2x^2 - 3}{x^2 + x +2}$

Let $y = p \left(x\right) + q \left(x\right)$ where $p \left(x\right) = x^{\cot x}$ and $q\left(x\right) = \dfrac{2x^2 -3}{x^2 + x + 2}$

Then $\dfrac{dy}{dx} = \dfrac{dp}{dx} + \dfrac{dq}{dx}$

$ \begin{aligned} p & = x^{\cot x} \\ \therefore \log p & = \log \left(x^{\cot x}\right) \\ \text{i.e. } \log p & = \cot x \log x \\ \therefore \dfrac{1}{p} \dfrac{dp}{dx} & = \dfrac{\cot x}{x} - \text{cosec}^2 x \log x \\ \text{i.e. } \dfrac{dp}{dx} & = p \left(\dfrac{\cot x}{x} - \text{cosec}^2 x \log x\right) \\ \text{i.e. } \dfrac{dp}{dx} & = x^{\cot x} \left(\dfrac{\cot x}{x} - \text{cosec}^2 x \log x\right) \end{aligned} $

$ \begin{aligned} q & = \dfrac{2x^2 - 3}{x^2 + x + 2} \\ \therefore \dfrac{dq}{dx} & = \dfrac{\left(x^2+x+2\right)\times \dfrac{d}{dx} \left(2x^2 -3\right) - \left(2x^2 -3\right) \times \dfrac{d}{dx} \left(x^2+x+2\right)}{\left(x^2+x+2\right)^2} \\ & = \dfrac{\left(x^2+x+2\right)\left(4x\right) - \left(2x^2 - 3\right)\left(2x+1\right)}{\left(x^2+x+2\right)^2} \\ & = \dfrac{4x^3 + 4x^2 + 8x - 4x^3 + 6x -2x^2 + 3}{\left(x^2 + x + 2\right)^2} \\ & = \dfrac{2x^2 + 14x + 3}{\left(x^2 + x + 2\right)^2} \end{aligned} $

$\therefore$ $\dfrac{dy}{dx} = \dfrac{dp}{dx} + \dfrac{dq}{dx} = x^{\cot x} \left(\dfrac{\cot x}{x} - \text{cosec}^2 x \log x\right) + \dfrac{2x^2 + 14x + 3}{\left(x^2 + x + 2\right)^2}$

Differentiation

If $y = \sqrt{\dfrac{1-\sin 2x}{1+\sin 2x}}$ show that $\dfrac{dy}{dx}+\sec^2 \left(\dfrac{\pi}{4}-x\right)=0$

$ \begin{aligned} \dfrac{1-\sin 2x}{1+\sin 2x} & = \dfrac{\sin^2 x + \cos^2 x -2 \sin x \cos x}{\sin^2 x + \cos^2 x + 2 \sin x \cos x} \\ & = \dfrac{\left(\cos x - \sin x\right)^2}{\left(\cos x + \sin x\right)^2} \\ \therefore y = \sqrt{\dfrac{1-\sin 2x}{1+\sin 2x}} & = \dfrac{\cos x - \sin x}{\cos x + \sin x} \\ & = \dfrac{\left(\cos x - \sin x\right)^2}{\left(\cos x + \sin x\right) \left(\cos x - \sin x\right)} \\ & = \dfrac{\cos^2 x + \sin^2 x -2 \sin x \cos x}{\cos^2 x - \sin^2 x} \\ & = \dfrac{1-\sin 2x}{\cos 2x} \end{aligned} $

$ \begin{aligned} \therefore \dfrac{dy}{dx} & = \dfrac{\cos 2x \times \dfrac{d}{dx}\left(1-\sin 2x\right)-\left(1-\sin 2x\right)\times \dfrac{d}{dx}\cos 2x}{\cos^2 2x} \\ & = \dfrac{\cos 2x \times \left(-2 \cos 2x\right) - \left(1-\sin 2x\right) \times \left(-2 \sin 2x\right)}{\cos^2 2x} \\ & = \dfrac{-2 \cos^2 2x - 2 \sin^2 2x + 2 \sin 2x}{\cos^2 2x} \\ & = \dfrac{-2\left(\cos^2 2x + \sin^2 2x\right) + 2 \sin 2x}{\cos^2 2x} \\ & = \dfrac{-2 + 2 \sin 2x}{\cos^2 2x} \\ & = \dfrac{-2 \left(1-\sin 2x\right)}{1-\sin^2 2x} \\ & = \dfrac{-2 \left(1-\sin 2x\right)}{\left(1+\sin 2x\right)\left(1-\sin 2x\right)} \\ & = \dfrac{-2}{1+\sin 2x} \end{aligned} $

$ \begin{aligned} \text{Now, } \cos \left(\dfrac{\pi}{4}-x\right) & = \cos \left(\dfrac{\pi}{4}\right) \cos x + \sin \left(\dfrac{\pi}{4}\right) \sin x \\ & = \dfrac{1}{\sqrt{2}} \cos x + \dfrac{1}{\sqrt{2}} \sin x \\ & = \dfrac{1}{\sqrt{2}} \left(\cos x + \sin x\right) \\ \therefore \cos^2 \left(\dfrac{\pi}{4}-x\right) & = \dfrac{1}{2} \left(\cos x + \sin x\right)^2 \\ & = \dfrac{1}{2} \left(\cos^2 x + \sin^2 x +2 \sin x \cos x\right) \\ & = \dfrac{1+\sin 2x}{2} \\ \therefore \sec^2 \left(\dfrac{\pi}{4}-x\right) & = \dfrac{2}{1+\sin 2x} \end{aligned} $

$\therefore$ $\dfrac{dy}{dx} = \dfrac{-2}{1+\sin 2x} = - \sec^2 \left(\dfrac{\pi}{4}-x\right)$

i.e. $\dfrac{dy}{dx} + \sec^2 \left(\dfrac{\pi}{4} - x\right) = 0$

Differentiation

Differentiate $\tan^{-1}\left[\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right]$ w.r.t x

Let $y=\tan^{-1}\left[\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right]$

$ \begin{aligned} \text{Now, } \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} & = \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \times \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} \\ & \\ & = \dfrac{\left(\sqrt{1+x} - \sqrt{1-x}\right)^2}{\left(\sqrt{1+x}\right)^2 - \left(\sqrt{1-x}\right)^2} \\ & \\ & = \dfrac{1+x+1-x-2\sqrt{1-x^2}}{1+x-\left(1-x\right)} \\ & \\ & = \dfrac{2-2\sqrt{1-x^2}}{1+x-1+x} \\ & \\ & = \dfrac{2-2\sqrt{1-x^2}}{2x} \\ & \\ & = \dfrac{1-\sqrt{1-x^2}}{x} \end{aligned} $

Let $x = \sin \theta$

Then, $\sqrt{1-x^2} = \sqrt{1-\sin^2 \theta} = \cos \theta$

$ \begin{aligned} \therefore \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} & = \dfrac{1-\sqrt{1-x^2}}{x} \\ & \\ & = \dfrac{1-\cos \theta}{\sin \theta} \\ & \\ & = \dfrac{2 \sin^2 \left(\dfrac{\theta}{2}\right)}{2 \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right)} \\ & \\ & = \dfrac{\sin \left(\dfrac{\theta}{2}\right)}{\cos \left(\dfrac{\theta}{2}\right)} = \tan \left(\dfrac{\theta}{2}\right) \end{aligned} $

$\therefore$ $y=\tan^{-1}\left[\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right] = \tan^{-1}\left[\tan\left(\dfrac{\theta}{2}\right)\right] = \dfrac{\theta}{2}$

$\therefore$ $\dfrac{dy}{dx} = \dfrac{1}{2} \dfrac{d\theta}{dx}$

Now, $x = \sin \theta$ $\implies$ $\theta = \sin^{-1}x$

$\therefore$ $\dfrac{d\theta}{dx} = \dfrac{1}{\sqrt{1-x^2}}$

$\therefore$ $\dfrac{dy}{dx}= \dfrac{1}{2}\dfrac{d\theta}{dx} = \dfrac{1}{2\sqrt{1-x^2}}$

Differentiation

Differentiate $\left(x\right)^{\cos x} + \left(\cos x\right)^x$ w.r.t x

Let $y = p\left(x\right) + q\left(x\right)$ where $p\left(x\right) = \left(x\right)^{\cos x}$ and $q\left(x\right)=\left(\cos x\right)^x$

Then $\dfrac{dy}{dx} = \dfrac{dp}{dx} + \dfrac{dq}{dx}$

$p = \left(x\right)^{\cos x}$

$\therefore$ $\log p = \log \left(x\right)^{\cos x} = \cos x \log \left(x\right)$

$\therefore$ $\dfrac{1}{p} \dfrac{dp}{dx} = \dfrac{\cos x}{x} - \sin x \log \left(x\right)$

$\therefore$ $\dfrac{dp}{dx} = p \left[\dfrac{\cos x}{x} - \sin x \log \left(x\right)\right]$

i.e. $\dfrac{dp}{dx} = \left(x\right)^{\cos x} \left[\dfrac{\cos x}{x} - \sin x \log \left(x\right)\right]$

$q = \left(\cos x\right)^x$

$\therefore$ $\log q = \log \left(\cos x\right)^x = x \log \left(\cos x\right)$

$\therefore$ $\dfrac{1}{q} \dfrac{dq}{dx} = \dfrac{x}{\cos x} \times \left(- \sin x\right) + \log \left(\cos x\right)$

$\therefore$ $\dfrac{dq}{dx} = q \left[- x \tan x + \log \left(\cos x\right)\right]$

i.e. $\dfrac{dq}{dx} = \left(\cos x\right)^x \left[-x\tan x + \log \left(\cos x\right)\right]$

$\therefore$ $\dfrac{dy}{dx} = \left(x\right)^{\cos x} \left[\dfrac{\cos x}{x} - \sin x \log \left(x\right)\right] + \left(\cos x\right)^x \left[-x\tan x + \log \left(\cos x\right)\right]$

Differentiation

Differentiate the function $f\left(x\right)=\sin \left(x^2 + 1\right)$ with respect to x from first principles.

$\begin{aligned} f'\left(x\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\sin \left[\left(x+h\right)^2 + 1\right]-\sin \left(x^2+1\right)}{h} \\ & \left[\text{Note: }\sin C - \sin D = 2 \cos \left(\dfrac{C+D}{2}\right) \sin \left(\dfrac{C-D}{2}\right)\right] \\ & = \lim\limits_{h \to 0} \; \dfrac{2 \cos \left[\dfrac{\left(x+h\right)^2 + 1 + x^2 + 1}{2}\right] \sin \left[\dfrac{\left(x+h\right)^2 + 1 - x^2 - 1}{2}\right]}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{2 \cos \left(\dfrac{2x^2 + 2xh+ h^2 + 2}{2}\right) \sin \left(\dfrac{2hx+h^2}{2}\right)}{h} \\ & = 2 \lim\limits_{h \to 0} \cos \left(x^2 + xh + 1 + \dfrac{h}{2}\right) \times \lim\limits_{h \to 0} \; \dfrac{\sin \left(hx+\dfrac{h^2}{2}\right)}{h\left(x+\dfrac{h}{2}\right)} \times \lim\limits_{h \to 0} \; \left(x+\dfrac{h}{2}\right) \\ & = 2 \times \cos \left(x^2+1\right) \times 1 \times x \\ & = 2 x \cos \left(x^2+1\right) \end{aligned}$

Differentiation

Differentiate the function $f\left(x\right)=\sqrt{\cos 3x}$ with respect to x from first principles.

$\begin{aligned} f'\left(x\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\sqrt{\cos \left(3x+3h\right)}-\sqrt{\cos 3x}}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\left[\sqrt{\cos \left(3x+3h\right)}-\sqrt{\cos 3x}\right] \left[\sqrt{\cos \left(3x+3h\right)}+\sqrt{\cos 3x}\right]}{h \left[\sqrt{\cos \left(3x+3h\right)}+\sqrt{\cos 3x}\right]} \\ & = \lim\limits_{h \to 0} \; \dfrac{\cos \left(3x+3h\right)-\cos 3x}{h \left[\sqrt{\cos \left(3x+3h\right)}+\sqrt{\cos 3x}\right]} \\ & \left[\text{Note: } \cos C - \cos D = 2 \sin \left(\dfrac{C+D}{2}\right) \sin \left(\dfrac{D-C}{2}\right)\right] \\ & = \lim\limits_{h \to 0} \; \dfrac{2 \sin \left(\dfrac{3x+3h+3x}{2}\right)\sin \left(\dfrac{3x-3x-3h}{2}\right)}{h\left[\sqrt{\cos \left(3x+3h\right)}+\sqrt{\cos 3x}\right]} \\ & = \lim\limits_{h \to 0} \; \dfrac{2 \sin \left(3x+\dfrac{3h}{2}\right) \sin \left(-\dfrac{3h}{2}\right)}{h \left[\sqrt{\cos \left(3x+3h\right)}+ \sqrt{\cos 3x}\right]} \\ & = -2 \; \lim\limits_{h \to 0} \; \dfrac{1}{\sqrt{\cos \left(3x+3h\right)}+\sqrt{\cos 3x}} \times \lim\limits_{h \to 0} \; \dfrac{\sin \left(\dfrac{3h}{2}\right)}{\dfrac{3h}{2}} \times \dfrac{3}{2} \times \lim\limits_{h \to 0} \; \sin \left(3x + \dfrac{3h}{2}\right) \\ & = -2 \times \dfrac{1}{\sqrt{\cos 3x}+\sqrt{\cos 3x}} \times 1 \times \dfrac{3}{2} \times \sin \left(3x\right) \\ & = \dfrac{-3\sin \left(3x\right)}{2\sqrt{\cos 3x}} \end{aligned}$

Differentiation

Show that the function $f\left(x\right)= \begin{cases} \left|x-3\right|, & x \geq 1 \\ \dfrac{x^2}{4}-\dfrac{3x}{2}+\dfrac{13}{4}, & x <1 \end{cases}$
is continuous as well as differentiable at $x=1$.

$\begin{aligned} \text{Left Hand Derivative } = Lf'\left(1\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(1-h\right)-f\left(1\right)}{-h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\dfrac{\left(1-h\right)^2}{4}-\dfrac{3\left(1-h\right)}{2}+\dfrac{13}{4}-\left|1-3\right|}{-h} \\ & = \lim\limits_{h \to 0} \; \dfrac{1-2h+h^2-6+6h+13-8}{-4h} \\ & = \lim\limits_{h \to 0} \; \dfrac{h^2 +4h}{-4h} \\ & = \lim\limits_{h \to 0} \; \dfrac{h+4}{-4} \\ & = -1 \end{aligned}$

$\begin{aligned} \text{Right Hand Derivative} = Rf'\left(1\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(1+h\right)-f\left(1\right)}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\left|1+h-3\right|-\left|1-3\right|}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{2-h-2}{h} = -1 \\ & \left[\begin{aligned} \text{Note:} & \because h \to 0, \; \text{h is a very small number} \\ & \therefore \; h-2<0 \implies \left|h-2\right|=-\left(h-2\right) \end{aligned}\right] \end{aligned}$

Since $Lf'\left(1\right)=Rf'\left(1\right)=-1$, $f\left(x\right)$ is differentiable at $x=1$.

Since the function is differentiable at $x=1$, it is continuous at $x=1$.

Differentiation

Find the derivative of $\dfrac{x^2-6}{x}$ with respect to x from first principle of derivatives.

Let $f\left(x\right)=\dfrac{x^2-6}{x}$

$\begin{aligned} f'\left(x\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\dfrac{\left(x+h\right)^2 -6}{x+h}-\left(\dfrac{x^2-6}{x}\right)}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\dfrac{x^2+2hx+h^2-6}{x+h}-\dfrac{x^2-6}{x}}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{x^3 + 2hx^2 + h^2 x -6x -x^3 -hx^2 +6x +6h}{hx\left(x+h\right)} \\ & = \lim\limits_{h \to 0} \; \dfrac{hx^2 + h^2 x +6h}{hx\left(x+h\right)} \\ & = \lim\limits_{h \to 0} \; \dfrac{x^2+hx+6}{x\left(x+h\right)} \\ & = \dfrac{x^2+0+6}{x\left(x+0\right)} \\ & = \dfrac{x^2+6}{x^2} \end{aligned}$

Differentiation

$f\left(x\right) = x^5 \text{sgn} \; x$, where $\text{sgn} \; x = \begin{cases} \dfrac{\left|x\right|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ $\;\;$ ($\text{sgn} \; x$ is the signum function) Examine the continuity and differentiability of $f\left(x\right)$ at $x=0$.

$\begin{aligned} \left|x\right| = \begin{cases} x, & x > 0 \\ -x, & x < 0 \end{cases} \end{aligned}$

$\begin{aligned} \therefore \; \text{sgn} \; x = \begin{cases} \dfrac{x}{x}=1, & x > 0 \\ & \\ -\dfrac{x}{x}=-1, & x < 0 \\ & \\ 0, & x=0 \end{cases} \end{aligned}$

$\begin{aligned} \therefore \; f\left(x\right) = x^5 \text{sgn} \; x = \begin{cases} x^5, & x > 0 \\ -x^5, & x < 0 \\ 0, & x=0 \end{cases} \end{aligned}$

$\begin{aligned} \text{Left Hand Derivative } \left(\text{at }x=0\right) = Lf'\left(0\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(0-h\right)-f\left(0\right)}{-h} \\ & = \lim\limits_{h \to 0} \; \dfrac{-\left(-h\right)^5 - 0}{-h} \\ & = \lim\limits_{h \to 0} \; \left(-h^4\right) = 0 \end{aligned}$

$\begin{aligned} \text{Right Hand Derivative } \left(\text{at }x=0\right) = Rf'\left(0\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(0+h\right)-f\left(0\right)}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{h^5 - 0}{h} \\ & = \lim\limits_{h \to 0} \; \left(h^4\right) = 0 \end{aligned}$

$\therefore$ $\;$ $Lf'\left(0\right) = Rf'\left(0\right) = 0$

$\implies$ $f\left(x\right)$ is differentiable at $x=0$

$\because$ $\;$ $f\left(x\right)$ is differentiable at $x=0$, $f\left(x\right)$ is also continuous at $x=0$.

Differentiation

Differentiate the function $f\left(x\right)=\cot \sqrt{x}$ from first principles.

$\begin{aligned} f'\left(x\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\cot \left(\sqrt{x+h}\right)-\cot \sqrt{x}}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\dfrac{\cos \left(\sqrt{x+h}\right)}{\sin \left(\sqrt{x+h}\right)}-\dfrac{\cos \sqrt{x}}{\sin \sqrt{x}}}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\sin \sqrt{x}\cos \left(\sqrt{x+h}\right)-\cos \sqrt{x}\sin \left(\sqrt{x+h}\right)}{\sin \sqrt{x} \sin \left(\sqrt{x+h}\right)h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\sin \left(\sqrt{x}-\sqrt{x+h}\right)}{\sin \sqrt{x} \sin \left(\sqrt{x+h}\right)h} \;\; \left[\text{Note: }\sin \left(\alpha - \beta\right)= \sin \alpha \cos \beta - \cos \alpha \sin \beta\right] \\ & = \dfrac{1}{\sin \sqrt{x}} \times \lim\limits_{h \to 0} \; \dfrac{1}{\sin \left(\sqrt{x+h}\right)} \times \lim\limits_{h \to 0} \; \dfrac{\sin \left(\sqrt{x}-\sqrt{x+h}\right)}{\sqrt{x}-\sqrt{x+h}} \times \lim\limits_{h \to 0} \; \dfrac{\sqrt{x}-\sqrt{x+h}}{h} \\ & = \dfrac{1}{\sin \sqrt{x}} \times \dfrac{1}{\sin \sqrt{x}} \times 1 \times \lim\limits_{h \to 0} \; \dfrac{\left(\sqrt{x}-\sqrt{x+h}\right)\left(\sqrt{x}+\sqrt{x+h}\right)}{h\left(\sqrt{x}+\sqrt{x+h}\right)} \\ & = \dfrac{1}{\sin^2 \sqrt{x}} \times \lim\limits_{h \to 0} \; \dfrac{x-x-h}{h\left(\sqrt{x}+\sqrt{x+h}\right)} \\ & = \text{cosec}^2 \sqrt{x} \times \lim\limits_{h \to 0} \; \dfrac{-1}{\sqrt{x}+\sqrt{x+h}} \\ & = \dfrac{-\text{cosec}^2 \sqrt{x}}{2\sqrt{x}} \end{aligned}$

Differentiation

For what choice of a and b, is the function $f\left(x\right) = \begin{cases} x^2, & x \leq c \\ ax+b, & x >c \end{cases}$ $\;\;$ differentiable at $x=c$?

Since the function $f\left(x\right)$ is differentiable at $x=c$ $\implies$ $f\left(x\right)$ is continuous at $x=c$

$\therefore$ $\;$ $\lim\limits_{x \to c^-} f\left(x\right) = \lim\limits_{x \to c^+} f\left(x\right) = f\left(c\right)$

i.e. $\lim\limits_{x \to c^-} x^2 = \lim\limits_{x \to c^+} ax+b = c^2$

$\implies$ $c^2 = ac+b \;\; \cdots$ (1)

Also, $f\left(x\right)$ is differentiable at $x=c$ $\implies$ $Lf'\left(c\right) = Rf'\left(c\right)$

$\therefore$ $\;$ $\lim\limits_{h \to 0} \dfrac{f\left(c-h\right)-f\left(c\right)}{-h} = \lim\limits_{h \to 0} \dfrac{f\left(c+h\right)-f\left(c\right)}{h}$

i.e. $\lim\limits_{h \to 0} \dfrac{\left(c-h\right)^2-c^2}{-h} = \lim\limits_{h \to 0} \dfrac{a\left(c+h\right)+b - c^2}{h}$

i.e. $\lim\limits_{h \to 0} \dfrac{c^2 -2ch + h^2 - c^2}{-h} = \lim\limits_{h \to 0} \dfrac{ac + ah + b - c^2}{h}$

i.e. $\lim\limits_{h \to 0} \left(2c-h\right) = \lim\limits_{h \to 0} \dfrac{c^2+ah-c^2}{h} \;\; \left[\text{From equation (1)}\right]$

i.e. $\lim\limits_{h \to 0} \left(2c-h\right) = \lim\limits_{h \to 0} a$

i.e. $2c=a$

Substituting the value of a in equation (1) gives

$c^2 = 2c^2 +b$ $\implies$ $b = -c^2$

Differentiation

Find the derivative of $\sin x + \cos x$ by first principle.

Let $f\left(x\right)=\sin x + \cos x$

$\begin{aligned} f'\left(x\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ & \\ & = \lim\limits_{h \to 0} \; \dfrac{\sin \left(x+h\right)+\cos \left(x+h\right)-\left(\sin x + \cos x\right)}{h} \\ & \\ & \left[\begin{aligned} \text{Note: } &\sin \left(\alpha + \beta\right)= \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ & \cos \left(\alpha + \beta\right)= \cos \alpha \cos \beta - \sin \alpha \sin \beta \end{aligned}\right] \\ & \\ & = \lim\limits_{h \to 0} \; \dfrac{\sin x \cos h+ \cos x \sin h + \cos x \cos h - \sin x \sin h - \left(\sin x + \cos x\right)}{h} \\ & \\ & = \lim\limits_{h \to 0} \; \dfrac{\cos h \left(\sin x + \cos x\right)-\left(\sin x + \cos x\right)+\sin h \left(\cos x - \sin x\right)}{h} \\ & \\ & = \lim\limits_{h \to 0} \; \dfrac{\left(\sin x +\cos x\right)\left(\cos h -1\right)+ \sin h \left(\cos x - \sin x\right)}{h} \\ & \\ & = \lim\limits_{h \to 0} \; \left\{\dfrac{\left(\sin x + \cos x\right) \left[-2 \sin^2 \left(\dfrac{h}{2}\right)\right]}{h} + \dfrac{\sin h \left(\cos x - \sin x\right)}{h} \right\} \\ & \\ & \left[\text{Note: }\cos \theta -1 = -2 \sin^2 \left(\dfrac{\theta}{2}\right)\right] \\ & \\ & = -2 \left(\sin x + \cos x\right) \; \lim\limits_{h \to 0} \; \left\{\left[\dfrac{\sin \left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}\right]^2 \times \dfrac{h^2}{4} \right\} + \left(\cos x - \sin x\right) \lim\limits_{h \to 0} \dfrac{\sin h}{h} \\ & \\ & = \dfrac{-1}{2} \left(\sin x + \cos x\right) \lim\limits_{\frac{h}{2} \to 0} \; \left[\dfrac{\sin \left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}\right]^2 \times \lim\limits_{h \to 0} h^2 + \left(\cos x - \sin x\right) \times 1 \\ & \\ & = \dfrac{-1}{2} \left(\sin x + \cos x\right) \times 1 \times 0 + \cos x - \sin x \\ & \\ & = \cos x - \sin x \end{aligned}$

Differentiation

Examine the continuity and differentiability of the function $f\left(x\right) = \begin{cases} xe^{-\left(\frac{1}{\left|x\right|}+\frac{1}{x}\right)}, & x \neq 0 \\ 0, & x = 0 \end{cases}$

$\begin{aligned} \left|x\right| = \begin{cases} x, & x > 0 \\ -x, & x < 0 \end{cases} \end{aligned}$

$\begin{aligned} \therefore \; \dfrac{1}{\left|x\right|} + \dfrac{1}{x} = \begin{cases} \dfrac{1}{x} + \dfrac{1}{x} = \dfrac{2}{x}, & x > 0 \\\\ \dfrac{-1}{x} + \dfrac{1}{x} = 0, & x < 0 \end{cases} \end{aligned}$

$\begin{aligned} \therefore \; f\left(x\right) = \begin{cases} xe^{-\frac{2}{x}}, & x > 0 \\ xe^{-0} = x, & x < 0 \\ 0, & x = 0 \end{cases} \end{aligned}$

$\begin{aligned} \text{Left Hand Derivative} = Lf'\left(0\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(0-h\right)-f\left(0\right)}{-h} \\ & = \lim\limits_{h \to 0} \; \dfrac{-h-0}{-h} \\ & = 1 \end{aligned}$

$\begin{aligned} \text{Right Hand Derivative} = Rf'\left(0\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(0+h\right)-f\left(0\right)}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{he^{-\frac{2}{h}}-0}{h} \\ & = \lim\limits_{h \to 0} \; e^{-\frac{2}{h}} \\ & = 0 \end{aligned}$

$\because$ $Lf'\left(0\right) \neq Rf'\left(0\right)$, the function is not differentiable at $x=0$.

$\because$ $Lf'\left(0\right)$ and $Rf'\left(0\right)$ are finite, $f\left(x\right)$ is continuous at $x=0$.

Differentiation

Differentiate by first principle $\sqrt{ax+b}$

Let $f\left(x\right)=\sqrt{ax+b}$

$\begin{aligned} f'\left(x\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\sqrt{a\left(x+h\right)+b}-\sqrt{ax+b}}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\left[\sqrt{a\left(x+h\right)+b}-\sqrt{ax+b} \right] \left[\sqrt{a\left(x+h\right)+b}+\sqrt{ax+b}\right]}{h \left[\sqrt{a\left(x+h\right)+b}+\sqrt{ax+b}\right]} \\ & = \lim\limits_{h \to 0} \; \dfrac{a\left(x+h\right)+b-ax-b}{h \left[\sqrt{a\left(x+h\right)+b}+\sqrt{ax+b}\right]} \\ & = \lim\limits_{h \to 0} \; \dfrac{ah}{h \left[\sqrt{a\left(x+h\right)+b}+\sqrt{ax+b}\right]} \\ & = a \; \lim\limits_{h \to 0} \; \dfrac{1}{\sqrt{a\left(x+h\right)+b}+\sqrt{ax+b}} \\ & = \dfrac{a}{\sqrt{ax+b}+\sqrt{ax+b}} \\ & = \dfrac{a}{2\sqrt{ax+b}} \end{aligned}$

Differentiation

Let $f\left(x\right) = \begin{cases} 0, & x < 0 \\ x^2, & x \geq 0 \end{cases}$ $\;\;\;$ Check the differentiability of $f\left(x\right)$ at $x=0$.

$\begin{aligned} \text{Left Hand Derivative} = Lf'\left(0\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(0-h\right)-f\left(0\right)}{-h} \\ & = \lim\limits_{h \to 0} \; \dfrac{0-0}{-h} \\ & = 0 \end{aligned}$

$\begin{aligned} \text{Right Hand Derivative} = Rf'\left(0\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(0+h\right)-f\left(0\right)}{h} \\ & = \lim\limits_{h \to 0} \; \dfrac{\left(0+h\right)^2-0}{h} \\ & = \lim\limits_{h \to 0} \; h \\ & = 0 \end{aligned}$

Since $Lf'\left(0\right) = Rf'\left(0\right)$, $f\left(x\right)$ is differentiable at $x=0$.

Differentiation

Find the derivative of $x \sin x$ with respect to x from first principle of derivatives.

Let $f\left(x\right)=x \sin x$

$\begin{aligned} f'\left(x\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\ & \\ & = \lim\limits_{h \to 0} \; \dfrac{\left(x+h\right)\sin \left(x+h\right)-x \sin x}{h} \\ & \\ & = \lim\limits_{h \to 0} \; \dfrac{x \sin \left(x+h\right)+h \sin \left(x+h\right)-x \sin x}{h} \\ & \\ & = \lim\limits_{h \to 0} \; \dfrac{x \left[\sin \left(x+h\right)-\sin x\right]}{h} + \lim\limits_{h \to 0} \; \dfrac{h \sin \left(x+h\right)}{h} \\ & \\ & \left[\text{Note: }\sin C - \sin D = 2 \cos \left(\dfrac{C+D}{2}\right)\sin \left(\dfrac{C-D}{2}\right)\right] \\ & \\ & = \lim\limits_{h \to 0} \; \dfrac{x \times 2 \cos \left(\dfrac{x+h+x}{2}\right) \sin \left(\dfrac{x+h-x}{2}\right)}{h} + \lim\limits_{h \to 0} \; \sin \left(x+h\right) \\ & \\ & = \lim\limits_{h \to 0} \; \dfrac{2x \cos \left(x+\dfrac{h}{2}\right) \sin \left(\dfrac{h}{2}\right)}{h} + \sin x \\ & \\ & = 2x \times \lim\limits_{h \to 0} \; \cos \left(x+\dfrac{h}{2}\right) \times \lim\limits_{\frac{h}{2} \to 0} \; \dfrac{\sin \left(\dfrac{h}{2}\right)}{\dfrac{h}{2}} \times \dfrac{1}{2} + \sin x \\ & \\ & \left[\text{Note: As }h \to 0, \; \dfrac{h}{2} \to 0 \right] \\ & \\ \therefore \; f'\left(x\right) & = x \times \cos x \times 1 + \sin x \\ & \\ & = x \cos x + \sin x \end{aligned}$

Continuity

If $f\left(x\right)=\begin{cases} \dfrac{\sin x - \cos x}{x - \frac{\pi}{4}}, & x \neq \dfrac{\pi}{4} \\ k, & x = \dfrac{\pi}{4} \end{cases}$ $\;\;$ is continuous at $x=\dfrac{\pi}{4}$, find k.

Since $f\left(x\right)$ is continuous at $x=\dfrac{\pi}{4}$ $\implies$ $f\left(\dfrac{\pi}{4}\right) = \lim\limits_{x \to \frac{\pi}{4}} f\left(x\right)$

i.e. $k = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\sin x - \cos x}{x - \frac{\pi}{4}}$

Let $x = \dfrac{\pi}{4}+h$

As $x \to \dfrac{\pi}{4}, \; h \to 0$

$\begin{aligned} \therefore \; k & = \lim\limits_{h \to 0} \; \dfrac{\sin \left(\dfrac{\pi}{4}+h\right)-\cos \left(\dfrac{\pi}{4}+h\right)}{h} \\ & \\ & \left[\begin{aligned} \text{Note: } & \sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ & \cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \end{aligned}\right] \\ & \\ & = \lim\limits_{h \to 0} \; \dfrac{\sin \left(\dfrac{\pi}{4}\right) \cos h + \cos \left(\dfrac{\pi}{4}\right)\sin h- \left[\cos \left(\dfrac{\pi}{4}\right) \cos h - \sin \left(\dfrac{\pi}{4} \right) \sin h\right]}{h} \\ & \\ & = \lim\limits_{h \to 0} \; \dfrac{\dfrac{\cos h}{\sqrt{2}} + \dfrac{\sin h}{\sqrt{2}} - \dfrac{\cos h}{\sqrt{2}}+ \dfrac{\sin h}{\sqrt{2}}}{h} \\ & \\ & = \dfrac{2}{\sqrt{2}} \; \lim\limits_{h \to 0} \; \dfrac{\sin h}{h} \\ & \\ & = \sqrt{2} \times 1 = \sqrt{2} \end{aligned}$

Continuity

If $f\left(x\right)= \begin{cases} \dfrac{\sin 3x}{7x} + \alpha, & x > 0 \\ x+3-\beta, & x < 0 \end{cases}$ $\;\;$ is continuous at $x=0$, find $\alpha + \beta$

Since $f\left(x\right)$ is continuous at $x=0$, $\lim\limits_{x \to 0^-} f\left(x\right) = \lim\limits_{x \to 0^+}f\left(x\right)$

i.e. $\lim\limits_{x \to 0^-} \; \left(x+3-\beta\right) = \lim\limits_{x \to 0^+} \; \left(\dfrac{\sin 3x}{7x}+\alpha\right)$

i.e. $3-\beta = \dfrac{3}{7} \; \lim\limits_{x \to 0^+} \; \dfrac{\sin 3x}{3x}+ \lim\limits_{x \to 0^+} \alpha$

i.e. $3-\beta = \dfrac{3}{7} \times 1 + \alpha$

i.e. $\alpha + \beta = 3-\dfrac{3}{7}=\dfrac{18}{7}$

Continuity

Let $f\left(x\right)= \begin{cases} x+a\sqrt{2} \sin x, & 0\leq x < \dfrac{\pi}{4} \\ & \\ 2x \cot x +b, & \dfrac{\pi}{4}\leq x < \dfrac{\pi}{2} \\ & \\ a \cos 2x - b \sin x, & \dfrac{\pi}{2}\leq x \leq \pi \end{cases}$

If $f\left(x\right)$ is continuous for $0 \leq x \leq \pi$, find a and b.

Since $f\left(x\right)$ is continuous for $0 \leq x \leq \pi$, therefore $f\left(x\right)$ must be continuous at $\dfrac{\pi}{4}$ and $\dfrac{\pi}{2}$

$\therefore$ $f\left(\dfrac{\pi}{4}\right) = \lim\limits_{x \to \frac{\pi}{4}^-} f\left(x\right) = \lim\limits_{x \to \frac{\pi}{4}^+} f\left(x\right)$ $\;\;$ $\cdots$ (1)

and $f\left(\dfrac{\pi}{2}\right) = \lim\limits_{x \to \frac{\pi}{2}^-} f\left(x\right) = \lim\limits_{x \to \frac{\pi}{2}^+} f\left(x\right)$ $\;\;$ $\cdots$ (2)

$\therefore$ Equation (1) gives

$2 \times \dfrac{\pi}{4} \times \cot \left(\dfrac{\pi}{4}\right)+b = \lim\limits_{x \to \frac{\pi}{4}^-} \; x+a\sqrt{2} \sin x = \lim\limits_{x \to \frac{\pi}{4}^+} \; 2x \cot x + b$

i.e. $\dfrac{\pi}{2}+b= \dfrac{\pi}{4}+a \sqrt{2} \sin \left(\dfrac{\pi}{4}\right)= 2 \times \dfrac{\pi}{4} \times \cot \left(\dfrac{\pi}{4}\right)+b$

i.e. $\dfrac{\pi}{2}+b= \dfrac{\pi}{4}+a\sqrt{2}\times \dfrac{1}{\sqrt{2}}=\dfrac{\pi}{2}+b$

i.e. $\dfrac{\pi}{4}+a = \dfrac{\pi}{2}+b$

i.e. $a-b=\dfrac{\pi}{4}$ $\cdots$ (3)

Equation (2) gives

$a \cos \left(2 \times \dfrac{\pi}{2}\right) - b\sin \left(\dfrac{\pi}{2}\right) = \lim\limits_{x \to \frac{\pi}{2}^-} \; 2x \cot x +b = \lim\limits_{x \to \frac{\pi}{2}^+} \; a \cos 2x - b\sin x$

i.e. $-a-b = 2 \times \dfrac{\pi}{2} \times \cot \left(\dfrac{\pi}{2}\right)+b = a \cos \left(2 \times \dfrac{\pi}{2}\right)- b\sin\left(\dfrac{\pi}{2}\right)$

i.e. $-a-b = b = -a-b$

i.e. $-a-b = b$

i.e. $a = -2b$ $\cdots$ (4)

Substituting the value of a from equation (4) in equation (3) gives

$-2b-b=\dfrac{\pi}{4}$ $\implies$ $b=-\dfrac{\pi}{12}$

$\therefore$ From equation (4), $a=2 \times \dfrac{\pi}{12}=\dfrac{\pi}{6}$

Continuity

Discuss the continuity of the function $f\left(x\right)= \begin{cases} \dfrac{x-\left|x\right|}{x}, & x \neq 0 \\ 2, & x=0 \end{cases}$ $\;$ at $x=0$

When $x>0, \; \left|x\right|=+x$

When $x<0, \; \left|x\right|=-x$

$\lim\limits_{x \to 0^+}f\left(x\right)=\lim\limits_{x \to 0} \; \dfrac{x-x}{x}=0$

$\lim\limits_{x \to 0^-}f\left(x\right)=\lim\limits_{x \to 0} \; \dfrac{x+x}{x}=\dfrac{2x}{x}=2$

$f\left(0\right)=2$

Since $\lim\limits_{x \to 0^-}f\left(x\right)=f\left(0\right) \neq \lim\limits_{x \to 0^+}f\left(x\right)$, function $f\left(x\right)$ is discontinuous at $x=0$.

Continuity

Find $\alpha$ and $\beta$ so that the function $f\left(x\right)$ defined by $f\left(x\right) = \begin{cases} -2 \sin x, & \text{for } -\pi \leq x \leq - \dfrac{\pi}{2} \\\\ \alpha \sin x + \beta, & \text{for } - \dfrac{\pi}{2} < x < \dfrac{\pi}{2} \\\\ \cos x, & \text{for } \dfrac{\pi}{2} \leq x \leq \pi \end{cases}$ $\;$ is continuous on $\left[-\pi,\pi\right]$

Since $f\left(x\right)$ is continuous on $\left[-\pi, \pi\right]$, therefore $f\left(x\right)$ must be continuous at $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$.

$\therefore$ $f\left(-\dfrac{\pi}{2}\right)= \lim\limits_{x \to -\frac{\pi}{2}^-} f\left(x\right) = \lim\limits_{x \to -\frac{\pi}{2}^+} f\left(x\right)$ $\cdots$ (1)

and $f\left(\dfrac{\pi}{2}\right) = \lim\limits_{x \to \frac{\pi}{2}^-}f\left(x\right)= \lim\limits_{x \to \frac{\pi}{2}^+}f\left(x\right)$ $\cdots$ (2)

$\therefore$ From equation (1) we have

$-2 \sin \left(-\dfrac{\pi}{2}\right) = \lim\limits_{x \to -\frac{\pi}{2}^-} \; -2\sin x = \lim\limits_{x \to -\frac{\pi}{2}^+} \; \alpha \sin x + \beta$

i.e. $-2 \sin \left(-\dfrac{\pi}{2}\right) = -2 \sin \left(-\dfrac{\pi}{2}\right) = \alpha \sin \left(-\dfrac{\pi}{2}\right)+ \beta$

i.e. $2 \sin \left(\dfrac{\pi}{2}\right) = 2 \sin \left(\dfrac{\pi}{2}\right) = -\alpha \sin \left(\dfrac{\pi}{2}\right)+\beta$ $\;\;\; \left[\text{Note: }\sin \left(-\theta\right)=-\sin \theta\right]$

i.e. $-\alpha + \beta = 2 \; \cdots (3)$

From equation (2) we have

$\cos \left(\dfrac{\pi}{2}\right) = \lim\limits_{x \to \frac{\pi}{2}^-} \; \alpha \sin x + \beta = \lim\limits_{x \to \frac{\pi}{2}^+} \; \cos x$

i.e. $\cos \left(\dfrac{\pi}{2}\right) = \alpha \sin \left(\dfrac{\pi}{2}\right)+\beta = \cos \left(\dfrac{\pi}{2}\right)$

i.e. $\alpha + \beta = 0 $

$\implies$ $\alpha = -\beta \; \cdots (4)$

Substituting the value of $\alpha$ in equation (3) gives

$2 \beta = 2$ $\implies$ $\beta = 1$

Substituting the value of $\beta$ in equation (4) gives

$\alpha = -1$

Continuity

If $f\left(x\right)= \dfrac{1-\tan x}{1-\sqrt{2} \sin x}$ for $x \neq \dfrac{\pi}{4}$, is continuous at $x=\dfrac{\pi}{4}$, find $f\left(\dfrac{\pi}{4}\right)$

$\begin{aligned} \lim\limits_{x \to \frac{\pi}{4}} f\left(x\right) & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{1-\tan x}{1-\sqrt{2}\sin x} \\ & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{1-\dfrac{\sin x}{\cos x}}{1-\sqrt{2}\sin x} \\ & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\cos x - \sin x}{\cos x \left(1-\sqrt{2}\sin x\right)} \\ & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\left(\cos x - \sin x\right)\left(cos x + \sin x\right)\left(1+\sqrt{2}\sin x\right)}{\cos x \left(1-\sqrt{2}\sin x\right) \left(1+\sqrt{2}\sin x\right) \left(\cos x + \sin x\right)} \\ & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\left(cos^2 x - \sin ^2 x\right)\left(1+\sqrt{2}\sin x\right)}{\cos x \left(1-2\sin^2 x\right)\left(\cos x + \sin x\right)} \\ & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{1+\sqrt{2}\sin x}{\cos x \left(\cos x + \sin x\right)}\\ & \\ & \left[\text{Note: }\cos 2\theta = \cos ^2 \theta - \sin^2 \theta = 1-2\sin^2 \theta\right] \\ & \\ & = \dfrac{1+\sqrt{2}\sin \left(\dfrac{\pi}{4}\right)}{\cos \left(\dfrac{\pi}{4}\right)\left[\cos \left(\dfrac{\pi}{4}\right)+ \sin \left(\dfrac{\pi}{4}\right)\right]} \\ & \\ & = \dfrac{1+\sqrt{2}\times \dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}\times \left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\right)} = 2 \end{aligned}$

Since $f\left(x\right)$ is continuous at $x=\dfrac{\pi}{4}$,

$\implies$ $f\left(\dfrac{\pi}{4}\right) = \lim\limits_{x \to \frac{\pi}{4}} f\left(x\right) = 2$

Continuity

Let $f\left(x\right)= \begin{cases} \dfrac{1-\sin^3 x}{3 \cos^2 x} & \text{if } x < \dfrac{\pi}{2} \\ & \\ a & \text{if } x = \dfrac{\pi}{2} \\ & \\ \dfrac{b\left(1-\sin x\right)}{\left(\pi - 2x\right)^2} & \text{if } x > \dfrac{\pi}{2} \end{cases}$
If $f\left(x\right)$ be a continuous function at $x=\dfrac{\pi}{2}$, find a and b.

$\begin{aligned} \lim\limits_{x \to \frac{\pi}{2}^-} f\left(x\right) & = \lim\limits_{x \to \frac{\pi}{2}^-} \; \dfrac{1-\sin^3 x}{3\cos^2 x} \\ & = \lim\limits_{x \to \frac{\pi}{2}^-} \; \dfrac{\left(1-\sin x\right)\left(1+\sin x + \sin^2 x\right)}{3 \left(1-\sin^2 x\right)} \\ & = \lim\limits_{x \to \frac{\pi}{2}^-} \; \dfrac{\left(1-\sin x\right)\left(1+\sin x + \sin^2 x\right)}{3 \left(1+\sin x\right)\left(1-\sin x\right)} \\ & = \lim\limits_{x \to \frac{\pi}{2}^-} \; \dfrac{1+\sin x + \sin^2 x}{3 \left(1+\sin x\right)} \\ & = \dfrac{1+1+1^2}{3\left(1+1\right)} = \dfrac{1}{2} \end{aligned}$

$\lim\limits_{x \to \frac{\pi}{2}^+} f\left(x\right) = \lim\limits_{x \to \frac{\pi}{2}^+} \; \dfrac{b\left(1-\sin x\right)}{\left(\pi - 2x\right)^2}$

Let $x = \dfrac{\pi}{2}+h$

Then, as $x \to \dfrac{\pi}{2}, \; h \to 0$

$\begin{aligned} \therefore \; \lim\limits_{x \to \frac{\pi}{2}^+}f\left(x\right) & = b \;\lim\limits_{h \to 0} \; \dfrac{1-\sin \left(\dfrac{\pi}{2}+h\right)}{\left[\pi - 2\left(\dfrac{\pi}{2}+h\right)\right]^2} \\ & = b\; \lim\limits_{h \to 0} \; \dfrac{1-\cos h}{4h^2} \;\; \left[\text{Note: }\sin \left(\dfrac{\pi}{2}+\theta\right)=\cos \theta\right] \\ & = b \;\lim\limits_{h \to 0} \; \dfrac{2 \sin^2 \left(\dfrac{h}{2}\right)}{4h^2} \;\; \left[\text{Note: }1-\cos 2 \theta = 2 \sin^2 \theta\right] \\ & = \dfrac{b}{2} \; \lim\limits_{h \to 0} \; \left[\dfrac{\sin \left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}\right]^2 \times \dfrac{1}{4} \\ & = \dfrac{b}{2} \times 1^2 \times \dfrac{1}{4} = \dfrac{b}{8} \end{aligned}$

Since $f\left(x\right)$ is continuous at $x=\dfrac{\pi}{2}$

$\implies$ $\lim\limits_{x \to \frac{\pi}{2}^-}f\left(x\right)=f\left(\dfrac{\pi}{2}\right)=\lim\limits_{x \to \frac{\pi}{2}^+}f\left(x\right)$

i.e. $\dfrac{1}{2} = a = \dfrac{b}{8}$

$\implies$ $a=\dfrac{1}{2}$ and $b=4$

Continuity

If the function $f\left(x\right) = \dfrac{\log_e \left(1+x\right)-\log_e \left(1-x\right)}{x}$ is continuous at $x=0$, find $f\left(0\right)$.

$\left[\begin{aligned} \text{Note: } & \log_e \left(1+x\right) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots \infty \\ & \log_e \left(1-x\right) = -x - \dfrac{x^2}{2} -\dfrac{x^3}{3} - \dfrac{x^4}{4} - \cdots \infty \end{aligned}\right]$

$\lim\limits_{x \to 0} f\left(x\right)$
$=\lim\limits_{x \to 0} \; \dfrac{\log_e \left(1+x\right)-\log_e \left(1-x\right)}{x}$
$\lim\limits_{x \to 0} \; \dfrac{\left[\left(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+ \cdots \infty\right)-\left(-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-\dfrac{x^4}{4}- \cdots \infty\right)\right]}{x}$
$= \lim\limits_{x \to 0} \; \dfrac{2x + \dfrac{2x^3}{3}+\dfrac{2x^5}{5}+ \cdots \infty}{x}$
$= \lim\limits_{x \to 0} \; \dfrac{2x \left(1+\dfrac{x^2}{3}+\dfrac{x^4}{5}+ \cdots \infty\right)}{x}$
$= 2 \lim\limits_{x \to 0} \; \left(1+\dfrac{x^2}{3}+\dfrac{x^4}{5}+ \cdots \infty\right)$
$= 2 \times 1 = 2$

Since the function $f\left(x\right)$ is continuous at $x=0$, $\lim\limits_{x \to 0} f\left(x\right)=f\left(0\right)$

$\therefore$ $f\left(0\right)=2$

Continuity

Locate the points of discontinuity, if any, for the function $f\left(x\right)= \begin{cases} \dfrac{x^3-64}{x^2-16}, & x \neq 4 \\ 1, & x=4 \end{cases}$

$\begin{aligned} f\left(x\right) & = \begin{cases} \dfrac{x^3-64}{x^2-16}, & x \neq 4 \\ 1, & x=4 \end{cases} \\ & \\ & = \begin{cases} \dfrac{\left(x-4\right)\left(x^2+4x+16\right)}{\left(x+4\right)\left(x-4\right)}, & x \neq 4 \\ 1, & x=4 \end{cases} \\ & \\ & = \begin{cases} \dfrac{x^2+4x+16}{x+4}, & x \neq 4 \\ 1, & x=4 \end{cases} \\ & \\ & = \begin{cases} \dfrac{x^2+4x+16}{x+4}, & x>4 \\ \dfrac{x^2+4x+16}{x+4}, & x<4 \\ 1, & x=4 \end{cases} \end{aligned}$

When $x>4$, $f\left(x\right)=\dfrac{x^2+4x+16}{x+4}$ is continuous (polynomial function).

When $x<4$, $f\left(x\right)=\dfrac{x^2+4x+16}{x+4}$ is continuous (polynomial function).

When $x=4$,

$\text{Left Hand Limit (LHL)} = \lim\limits_{x \to 4^-}f\left(x\right)=\lim\limits_{x \to 4^-} \; \dfrac{x^2+4x+16}{x+4}$

Let $x=4-h$

As $x \to 4^-, \; h \to 0$

$\begin{aligned} \therefore \text{LHL} & =\lim\limits_{h \to 0} \; \dfrac{\left(4-h\right)^2+4\left(4-h\right)+16}{4-h+4} \\ & = \lim\limits_{h \to 0} \; \dfrac{16-8h+h^2+16-4h+16}{8-h} \\ & = \lim\limits_{h \to 0} \; \dfrac{h^2-12h+48}{8-h} \\ & = \dfrac{0-0+48}{8-0} = 6 \end{aligned}$

$\text{Right Hand Limit (RHL)} = \lim\limits_{x \to 4^+}f\left(x\right)=\lim\limits_{x \to 4^+} \; \dfrac{x^2+4x+16}{x+4}$

Let $x=4+h$

As $x \to 4^+, \; h \to 0$

$\begin{aligned} \therefore \text{RHL} & =\lim\limits_{h \to 0} \; \dfrac{\left(4+h\right)^2+4\left(4+h\right)+16}{4+h+4} \\ & = \lim\limits_{h \to 0} \; \dfrac{16+8h+h^2+16+4h+16}{8+h} \\ & = \lim\limits_{h \to 0} \; \dfrac{h^2+12h+48}{8+h} \\ & = \dfrac{0+0+48}{8+0} = 6 \end{aligned}$

$f\left(4\right)=1$

$\therefore$ $\text{LHL}=\text{RHL}\neq f\left(4\right)$

$\implies$ $x=4$ is a point of discontinuity.

Continuity

Find the value of constant k so that the function $f\left(x\right)= \begin{cases} \dfrac{k \cos x}{\pi - 2x}, & \text{if } x \neq \dfrac{\pi}{2} \\\\ 3, & \text{if } x = \dfrac{\pi}{2} \end{cases}$
is continuous at $x=\dfrac{\pi}{2}$

$\lim\limits_{x \to \frac{\pi}{2}} f\left(x\right)=\lim\limits_{x \to \frac{\pi}{2}} \; \dfrac{k \cos x}{\pi - 2x}$

Let $x=\dfrac{\pi}{2}+p$

As $x \to \dfrac{\pi}{2}, \; p \to 0$

$\begin{aligned} \therefore \; \lim\limits_{x \to \frac{\pi}{2}} f\left(x\right) & = \lim\limits_{p \to 0} \; \dfrac{k \cos \left(\dfrac{\pi}{2}+p\right)}{\pi -2 \left(\dfrac{\pi}{2}+p\right)} \\ & = \lim\limits_{p \to 0} \; \dfrac{-k \sin p}{-2p} \;\; \left[\text{Note: }\cos \left(\dfrac{\pi}{2}+\theta\right)=-\sin \theta\right] \\ & = \dfrac{k}{2} \; \lim\limits_{p \to 0} \; \dfrac{\sin p}{p} \\ & = \dfrac{k}{2} \times 1 = \dfrac{k}{2} \;\; \cdots (1) \end{aligned}$

$f\left(\dfrac{\pi}{2}\right)=3$ $\;\;$ $\cdots$ (2)

Since $f\left(x\right)$ is continuous at $x=\dfrac{\pi}{2}$, $\;$ $\lim\limits_{x \to \frac{\pi}{2}}f\left(x\right)=f\left(\dfrac{\pi}{2}\right)$

i.e. $\dfrac{k}{2}=3$ [from equations (1) and (2)]

$\implies$ $k=6$

Continuity

The function $f\left(x\right)$ is defined as

$f\left(x\right)= \begin{cases} x^2 + ax + b, & 0 \leq x < 2 \\ & \\ 3 x + 2, & 2 \leq x \leq 4 \\ & \\ 2 a x + 5 b, & 4 < x \leq 8 \end{cases}$

is continuous in $\left[0,8\right]$. Find the values of a and b.

Since $f\left(x\right)$ is continuous in $\left[0,8\right]$, therefore $f\left(x\right)$ must be continuous at 2 and 4.

$\therefore$ $f\left(2\right)= \lim\limits_{x \to 2^-} f\left(x\right) = \lim\limits_{x \to 2^+} f\left(x\right)$ $\cdots$ (1)

and $f\left(4\right) = \lim\limits_{x \to 4^-}f\left(x\right)= \lim\limits_{x \to 4^+}f\left(x\right)$ $\cdots$ (2)

$\therefore$ From equation (1) we have

$\left(3\times2\right)+2 = \lim\limits_{x \to 2^-} \; x^2+ax+b = \lim\limits_{x \to 2^+} \; 3x+2$

i.e. $8=4+2a+b=8$

i.e. $2a+b=4$ $\cdots$ (3)

From equation (2) we have

$\left(3 \times 4\right)+2 = \lim\limits_{x \to 4^-} \; 3x+2 = \lim\limits_{x \to 4^+} \; 2ax+5b$

i.e. $14=14=8a+5b$

i.e. $8a+5b=14$ $\cdots$ (4)

Multiplying equation (3) with 5 and subtracting from equation (4) gives

$2a=6$ $\implies$ $a=3$

$\therefore$ From equation (3), $b=4-\left(2\times 3\right)=-2$

Continuity

For what value(s) of a and b, $f\left(x\right)= \begin{cases} \dfrac{x+2}{\left|x+2\right|}+a, & x<-2 \\ & \\ a+b, & x=-2 \\ & \\ \dfrac{x+2}{\left|x+2\right|}+b, & x>-2 \end{cases}$
is continuous at $x=-2$

$f\left(-2\right)=a+b$ $\cdots$ (1)

When $x<-2, \; \left|x+2\right|=-\left(x+2\right)$

When $x>-2, \; \left|x+2\right|=\left(x+2\right)$

$\begin{aligned} \therefore \lim\limits_{x \to -2^-} f\left(x\right) & =\lim\limits_{x\to -2^-} \; \dfrac{x+2}{\left|x+2\right|}+a \\ & \\ & = \lim\limits_{x \to -2} \; \dfrac{x+2}{-\left(x+2\right)}+a \\ & \\ & = -1+a \;\;\; \cdots (2) \end{aligned}$

$\begin{aligned} \lim\limits_{x \to -2^+} f\left(x\right) & =\lim\limits_{x\to -2^+} \; \dfrac{x+2}{\left|x+2\right|}+b \\ & \\ & = \lim\limits_{x \to -2} \; \dfrac{x+2}{\left(x+2\right)}+b \\ & \\ & = 1+b \;\;\; \cdots (3) \end{aligned}$

Since the function is continuous at $x=-2$,

$\lim\limits_{x \to -2^-}f\left(x\right)=f\left(-2\right)=\lim\limits_{x \to -2^+}f\left(x\right)$

i.e. $-1+a=a+b$ $\implies$ $b=-1$ [from equations (1) and (2)]

and $1+b=a+b$ $\implies$ $a=1$ [from equations (1) and (3)]

Limits

Evaluate $\lim\limits_{x \to \frac{\pi}{2}} \; \left(1+\cos x\right)^{3 \sec x}$

Let $\cos x = t$

Then as $x \to \dfrac{\pi}{2}, \; t \to 0$

$\begin{aligned} \lim\limits_{x \to \frac{\pi}{2}} \; \left(1+\cos x\right)^{3 \sec x} & = \lim\limits_{x \to \frac{\pi}{2}} \; \left(1+\cos x\right)^{\frac{3}{\cos x}} \\ & = \lim\limits_{t \to 0} \; \left(1+t\right)^{\frac{3}{t}} \\ & = \left\{\lim\limits_{t \to 0} \; \left(1+t\right)^{\frac{1}{t}}\right\}^3 \\ & = e^3 \;\; \left[\text{Note: }\lim\limits_{x \to 0} \left(1+x\right)^{\frac{1}{x}}=e\right] \end{aligned}$

Limits

Evaluate: $\lim\limits_{x \to \frac{\pi}{6}} \; \dfrac{2-\sqrt{3}\cos x - \sin x}{\left(6x-\pi\right)^2}$

Let $x-\dfrac{\pi}{6} = p$ $\implies$ $6x-\pi=6p \;$ and $\; x=\dfrac{\pi}{6}+p$

As $x \to \dfrac{\pi}{6}, \; p \to 0$

$\lim\limits_{x \to \frac{\pi}{6}} \; \dfrac{2-\sqrt{3}\cos x - \sin x}{\left(6x-\pi\right)^2}$

$= \lim\limits_{p \to 0} \; \dfrac{2-\sqrt{3}\cos\left(\dfrac{\pi}{6}+p\right)-\sin \left(\dfrac{\pi}{6}+p\right)}{\left(6p\right)^2}$

$= \lim\limits_{p \to 0} \; \dfrac{2-\sqrt{3}\left(\cos \dfrac{\pi}{6} \cos p - \sin \dfrac{\pi}{6} \sin p\right)-\left(\sin \dfrac{\pi}{6} \cos p + \cos \dfrac{\pi}{6} \sin p\right)}{36p^2}$

$\left[\text{Note: }\sin\left(\alpha + \beta\right)=\sin \alpha \cos \beta + \cos \alpha \sin \beta\right]$

$\left[\text{Note: }\cos \left(\alpha + \beta\right)=\cos \alpha \cos \beta - \sin \alpha \sin \beta\right]$

$= \lim\limits_{p \to 0} \; \dfrac{2-\sqrt{3}\left(\dfrac{\sqrt{3}}{2}\cos p - \dfrac{1}{2}\sin p\right)-\left(\dfrac{1}{2}\cos p+ \dfrac{\sqrt{3}}{2}\sin p\right)}{36p^2}$

$= \lim\limits_{p \to 0} \; \dfrac{2-\dfrac{3}{2}\cos p + \dfrac{\sqrt{3}}{2}\sin p - \dfrac{1}{2}\cos p - \dfrac{\sqrt{3}}{2}\sin p}{36p^2}$

$= \lim\limits_{p \to 0} \; \dfrac{2-2\cos p}{36p^2}$

$= \lim\limits_{p \to 0} \; \dfrac{2\left(1-\cos p\right)}{36p^2}$

$= \dfrac{1}{18} \; \lim\limits_{p \to 0} \; \dfrac{2 \sin^2 \left(\dfrac{p}{2}\right)}{p^2} \;\; \left[\text{Note: }1-\cos \theta = 2 \sin^2 \left(\dfrac{\theta}{2}\right)\right]$

$= \dfrac{1}{9} \; \lim\limits_{\frac{p}{2} \to 0} \; \left[\dfrac{\sin \left(\dfrac{p}{2}\right)}{\dfrac{p}{2}}\right]^2 \times \dfrac{1}{4} \;\; \left[\text{Note: As }p \to 0, \; \dfrac{p}{2} \to 0\right]$

$= \dfrac{1}{9} \times 1 \times \dfrac{1}{4} = \dfrac{1}{36}$

Limits

Evaluate: $\lim\limits_{x \to \frac{\pi}{3}} \; \dfrac{\sqrt{3}-\tan x}{\pi - 3x}$

Let $\dfrac{\pi}{3}-x=p$ $\implies$ $x=\dfrac{\pi}{3}-p$ and $\pi-3x=3p$

As $x \to \dfrac{\pi}{3}, \; p \to 0$

$\begin{aligned} \therefore \lim\limits_{x \to \frac{\pi}{3}} \; \dfrac{\sqrt{3}-\tan x}{\pi -3x} & = \lim\limits_{p \to 0} \; \dfrac{\sqrt{3}-\tan\left(\dfrac{\pi}{3}-p\right)}{3p} \\ & = \lim\limits_{p \to 0} \; \dfrac{\sqrt{3}-\left(\dfrac{\tan \dfrac{\pi}{3}-\tan p}{1+\tan \dfrac{\pi}{3}\times \tan p}\right)}{3p} \\ & \left[\text{Note: }\tan \left(\alpha-\beta\right)= \dfrac{\tan \alpha - \tan \beta}{1+\tan \alpha \tan \beta}\right] \\ & = \lim\limits_{p \to 0} \; \dfrac{\sqrt{3}-\left(\dfrac{\sqrt{3}-\tan p}{1+ \sqrt{3}\tan p}\right)}{3p} \\ & = \lim\limits_{p \to 0} \; \dfrac{\sqrt{3}+3 \tan p -\sqrt{3}+ \tan p}{3p \left(1+\sqrt{3}\tan p\right)} \\ & = \lim\limits_{p \to 0} \; \dfrac{4 \tan p}{3p \left(1+\sqrt{3}\tan p\right)} \\ & = \dfrac{4}{3} \; \lim\limits_{p \to 0} \; \dfrac{\tan p}{p} \times \dfrac{1}{\lim\limits_{p \to 0} \; \left(1+\sqrt{3}\tan p\right)} \\ & = \dfrac{4}{3} \times 1 \times \dfrac{1}{\left(1+0\right)} = \dfrac{4}{3} \end{aligned}$

Limits

Find $\lim\limits_{x \to 0} \; \dfrac{3x + \left|x\right|}{2x-\left|x\right|}$, if it exists.

When $x>0, \; \left|x\right|=+x$

When $x<0, \; \left|x\right|=-x$

$\begin{aligned} \therefore \lim\limits_{x \to 0} \; \dfrac{3x+\left|x\right|}{2x-\left|x\right|} & = \lim\limits_{x \to 0^+} \; \dfrac{3x+x}{2x-x} \;\; for \; x > 0 \\ & = \lim\limits_{x \to 0^+} \; \dfrac{4x}{x} \\ & = \lim\limits_{x \to 0^+} 4 = 4 \end{aligned}$

$\begin{aligned} \text{Also, } \lim\limits_{x \to 0} \; \dfrac{3x+\left|x\right|}{2x-\left|x\right|} & = \lim\limits_{x \to 0^-} \; \dfrac{3x-x}{2x+x} \;\; for \; x < 0 \\ & = \lim\limits_{x \to 0^-} \; \dfrac{2x}{3x} \\ & = \lim\limits_{x \to 0^-} \dfrac{2}{3} = \dfrac{2}{3} \end{aligned}$

$\therefore$ $\lim\limits_{x \to 0^+} \; \dfrac{3x+\left|x\right|}{2x-\left|x\right|} \neq \lim\limits_{x \to 0^-} \; \dfrac{3x+\left|x\right|}{2x-\left|x\right|}$

$\implies$ $\lim\limits_{x \to 0} \; \dfrac{3x+\left|x\right|}{2x-\left|x\right|}$ does not exist.

Limits

Evaluate: $\lim\limits_{x \to 0} \; \dfrac{\sin 2x + \sin 6x}{\sin 5x - \sin 3x}$

$\sin C + \sin D = 2 \sin \left(\dfrac{C+D}{2}\right) \cos \left(\dfrac{C-D}{2}\right)$
$\sin C - \sin D = 2 \cos \left(\dfrac{C+D}{2}\right) \sin \left(\dfrac{C-D}{2}\right)$
$\begin{aligned} \therefore \lim\limits_{x \to 0} \;\; \dfrac{\sin 2x + \sin 6x}{\sin 5x - \sin 3x} & = \lim\limits_{x \to 0} \dfrac{2 \sin \left(\dfrac{6x+2x}{2}\right) \cos \left(\dfrac{6x-2x}{2}\right)}{2 \cos \left(\dfrac{5x+3x}{2} \right) \sin \left(\dfrac{5x-3x}{2}\right)} \\ & \\ & = \lim\limits_{x \to 0} \; \dfrac{\sin 4x \cos 2x}{\cos 4x \sin x} \\ & \\ & = \dfrac{\lim\limits_{4x \to 0}\; \dfrac{\sin 4x}{4x}\times 4 \times \lim\limits_{x \to 0} \; \cos 2x}{\lim\limits_{x \to 0} \; \cos 4x \times \lim\limits_{x \to 0} \; \dfrac{\sin x}{x}} \\ & \\ & \left[\text{Note: Dividing numerator and denominator by x} \right]\\ & \\ & \left[\text{Note: As } x \to 0, 4x \to 0\right] \\ & \\ & = \dfrac{1 \times 4 \times 1}{1 \times 1} = 4 \end{aligned}$

Limits

Evaluate: $\lim\limits_{x \to \infty} \; \left[x-\sqrt{x^2+x}\right]$

Let $x = \dfrac{1}{y}$
Then, as $x \to \infty, \; y \to 0$
$\begin{aligned} \therefore \lim\limits_{x \to \infty} \; \left[x-\sqrt{x^2+x}\right] & = \lim\limits_{y \to 0} \; \left[\dfrac{1}{y}-\sqrt{\dfrac{1}{y^2}+\dfrac{1}{y}}\right] \\ & = \lim\limits_{y \to 0} \; \left[\dfrac{1}{y}-\dfrac{\sqrt{1+y}}{y}\right] \\ & = \lim\limits_{y \to 0} \; \left[\dfrac{1-\sqrt{1+y}}{y}\right] \\ & = \lim\limits_{y \to 0} \; \left[\dfrac{\left(1-\sqrt{1+y}\right) \left(1+\sqrt{1+y}\right)}{y \left(1+\sqrt{1+y}\right)}\right] \\ & = \lim\limits_{y \to 0} \; \dfrac{1-1-y}{y\left(1+\sqrt{1+y}\right)} \\ & = \lim\limits_{y \to 0} \; \dfrac{-1}{1+\sqrt{1+y}} \\ & = \dfrac{-1}{1+\sqrt{1+0}} = \dfrac{-1}{2} \end{aligned}$

Limits

If $\lim\limits_{x \to 1} \dfrac{x^4 -1}{x-1} = \lim\limits_{x \to k} \dfrac{x^3 - k^3}{x^2 - k^2}$, find the value of k.

$\begin{aligned} \lim\limits_{x \to 1} \dfrac{x^4 -1}{x-1} & = \lim\limits_{x \to 1} \dfrac{x^4 -1^4}{x-1} \\ & = 4 \times 1^{4-1} \;\; \left[\text{Note: } \lim\limits_{x \to a} \dfrac{x^n - a^n}{x-a} = na^{n-1}\right] \\ & = 4 \end{aligned}$
$\begin{aligned} \lim\limits_{x \to k} \dfrac{x^3 - k^3}{x^2 - k^2} & = \lim\limits_{x \to k} \; \dfrac{\dfrac{x^3 - k^3}{x-k}}{\dfrac{x^2 - k^2}{x-k}} \\ & = \dfrac{\lim\limits_{x \to k} \dfrac{x^3-k^3}{x-k}}{\lim\limits_{x \to k} \dfrac{x^2-k^2}{x-k}} = \dfrac{3k^{3-1}}{2k^{2-1}} = \dfrac{3k}{2} \end{aligned}$
$\therefore$ $\lim\limits_{x \to 1} \dfrac{x^4 -1}{x-1} = \lim\limits_{x \to k} \dfrac{x^3 - k^3}{x^2 - k^2}$ $\implies$ $4=\dfrac{3k}{2}$ $\implies$ $k = \dfrac{8}{3}$

Limits

Evaluate: $\lim\limits_{x \to 0} \dfrac{\tan 2x - \sin 2x}{x^3}$

$\begin{aligned} \lim\limits_{x \to 0} \dfrac{\tan 2x - \sin 2x}{x^3} & = \lim\limits_{x \to 0} \dfrac{\dfrac{\sin 2x}{\cos 2x}- \sin 2x}{x^3} \\ & = \lim\limits_{x \to 0} \dfrac{\sin 2x - \sin 2x \cos 2x}{x^3 \cos 2x} \\ & = \lim\limits_{x \to 0} \dfrac{\sin 2x \left(1-\cos 2x\right)}{x^3 \cos 2x} \\ & = \lim\limits_{x \to 0} \dfrac{\sin 2x \times 2 \sin^2 x}{x^3 \cos 2x} \;\; \left[\text{Note: }1- \cos 2\theta = 2 \sin^2 \theta\right] \\ & = 2 \lim\limits_{2x \to 0} \dfrac{\sin 2x}{2x} \times 2 \lim\limits_{x \to 0} \left(\dfrac{\sin x}{x}\right)^2 \times \lim\limits_{x \to 0} \dfrac{1}{\cos 2x} \\ & \left[\text{Note: As }x \to 0, \; 2x \to 0\right] \\ & = 2 \times 1 \times 2 \times 1^2 \times 1 \\ & = 4 \end{aligned}$

Limits

Evaluate: $\lim\limits_{x \to 0} \dfrac{\sqrt[3]{\left(1+x\right)}-\sqrt[3]{\left(1-x\right)}}{x}$

$\begin{aligned} \lim\limits_{x \to 0} \dfrac{\sqrt[3]{\left(1+x\right)}-\sqrt[3]{\left(1-x\right)}}{x} & = \lim\limits_{x \to 0} \dfrac{\left(1+x\right)^{1/3}-1 - \left(1-x\right)^{1/3}+1}{x} \\ & \\ & = \lim\limits_{x \to 0} \; \dfrac{\left(1+x\right)^{1/3}-1}{x} - \lim\limits_{x \to 0} \; \dfrac{\left(1-x\right)^{1/3}-1}{x} \\ & \\ & = L_1 - L_2 \end{aligned}$

Consider $L_1 = \lim\limits_{x \to 0} \dfrac{\left(1+x\right)^{1/3}-1}{x}$

Let $1+x=p$

Then as $x \to 0$, $p \to 1$

Also, $x = p-1$

$\therefore$ $L_1 = \lim\limits_{p \to 1} \dfrac{p^{1/3}-1}{p-1} = \dfrac{1}{3} \times 1^{\frac{1}{3}-1}=\dfrac{1}{3}$ $\left[\text{Note: }\lim\limits_{x \to a} \dfrac{x^n - a^n}{x-a}=na^{n-1}\right]$

Consider $L_2 = \lim\limits_{x \to 0} \dfrac{\left(1-x\right)^{1/3}-1}{x}$

Let $1-x=q$

Then, as $x \to 0$, $q \to 1$

Also, $x=1-q$

$\therefore$ $L_2 = \lim\limits_{q \to 1} \dfrac{q^{1/3}-1}{1-q} = - \lim\limits_{q \to 1} \dfrac{q^{1/3}-1}{q-1}=-\dfrac{1}{3} \times 1^{\frac{1}{3}-1} = - \dfrac{1}{3}$

$\therefore$ $\lim\limits_{x \to 0} \dfrac{\sqrt[3]{\left(1+x\right)}-\sqrt[3]{\left(1-x\right)}}{x} = L_1 - L_2 = \dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}$

Limits

Evaluate: $\lim\limits_{x \to 0} \; \dfrac{\cos 2x -1}{\cos x -1}$

$\begin{aligned} \lim\limits_{x \to 0} \; \dfrac{\cos 2x -1}{\cos x -1} & = \lim\limits_{x \to 0} \; \dfrac{-2\sin^2 x}{-2 \sin^2 \left(\dfrac{x}{2}\right)} \;\; \left[\text{Note: }\cos 2 \theta -1 = -2 \sin^2 \theta\right] \\ & \\ & = \lim\limits_{x \to 0} \dfrac{\sin^2 x}{\sin^2 \left(\dfrac{x}{2}\right)} \\ & \\ & = \lim\limits_{x \to 0} \dfrac{\left(\dfrac{\sin x}{x}\right)^2}{\left[\dfrac{\sin \left(\dfrac{x}{2}\right)}{\dfrac{x}{2}}\right]^2 \times \dfrac{1}{4}} \\ & \\ & = 4 \times \dfrac{1^2}{1^2} = 4 \end{aligned}$

Limits

Evaluate: $\lim\limits_{x \to \sqrt{3}} \;\; \dfrac{x^2 -3}{x^2+3\sqrt{3}x - 12}$

$\begin{aligned} \lim\limits_{x \to \sqrt{3}} \;\; \dfrac{x^2 - 3}{x^2+3\sqrt{3}x-12} & = \lim\limits_{x \to \sqrt{3}} \;\; \dfrac{\left(x+\sqrt{3}\right) \left(x-\sqrt{3}\right)}{x^2+4\sqrt{3}x - \sqrt{3}x -12} \\ & \\ & = \lim\limits_{x \to \sqrt{3}} \;\; \dfrac{\left(x+\sqrt{3}\right) \left(x-\sqrt{3}\right)}{x \left(x+4\sqrt{3}\right)-\sqrt{3} \left(x+4\sqrt{3}\right)} \\ & \\ & = \lim\limits_{x \to \sqrt{3}} \;\; \dfrac{\left(x+\sqrt{3}\right) \left(x-\sqrt{3}\right)}{\left(x+4\sqrt{3}\right) \left(x-\sqrt{3}\right)} \\ & \\ & = \lim\limits_{x \to \sqrt{3}} \;\; \dfrac{x+\sqrt{3}}{x+4\sqrt{3}} \\ & \\ & = \dfrac{\sqrt{3}+ \sqrt{3}}{\sqrt{3}+ 4\sqrt{3}} \\ & \\ & = \dfrac{2\sqrt{3}}{5\sqrt{3}} \\ & \\ & = \dfrac{2}{5} \end{aligned}$

Limits

Evaluate: $\lim\limits_{x \to 0} \dfrac{\sin x - \tan x}{x^3}$

$\begin{aligned} \lim\limits_{x \to 0} \dfrac{\sin x - \tan x}{x^3} & = \lim\limits_{x \to 0} \dfrac{\sin x - \dfrac{\sin x}{\cos x}}{x^3} \\ & \\ & = \lim\limits_{x \to 0} \dfrac{\sin x \cos x - \sin x}{x^3 \cos x} \\ & \\ & = \lim\limits_{x \to 0} \dfrac{\sin x \left(\cos x -1\right)}{x^3 \cos x} \\ & \\ & = \lim\limits_{x \to 0} \dfrac{-2 \sin x \sin^2 \left(\dfrac{x}{2}\right)}{x^3 \cos x} \; \left[\text{Note: }\cos \theta -1 = -2 \sin^2 \left(\dfrac{\theta}{2}\right)\right] \\ & \\ & = -2 \lim\limits_{x \to 0} \dfrac{\sin x}{x} \times \dfrac{1}{4} \lim\limits_{\frac{x}{2} \to 0} \left[\dfrac{\sin \left(\dfrac{x}{2}\right)}{\dfrac{x}{2}}\right]^2 \times \lim\limits_{x \to 0} \dfrac{1}{\cos x} \\ & \\ & \left[\text{Note: As }x \to 0, \dfrac{x}{2} \to 0\right] \\ & \\ & = -2 \times 1 \times \dfrac{1}{4} \times 1^2 \times 1 \\ & \\ & = \dfrac{-1}{2} \end{aligned}$

Quadratic Equations: Word Problems

Two years ago, a man's age was three times the square of his daughter's age. Three years hence his age will be four times his daughter's age. Find the present age of the man.

Let the daughter's present age $=x$ years

Then, two years ago, daughter's age $= \left(x-2\right)$ years

$\therefore$ Man's age two years ago $= 3 \left(x-2\right)^2 = 3 \left(x^2 - 4x + 4\right) = 3x^2-12x+12$ years

$\therefore$ Man's present age $= 3x^2 - 12x + 12 + 2 = 3x^2 -12x +14$ years

Man's age three years hence $= 3x^2 - 12x + 14 + 3 = 3x^2 -12x + 17$ years

Daughter's age three years hence $= x+3$ years

$\therefore$ As per sum,

$3x^2-12x+17=4\left(x+3\right)$

i.e. $3x^2-12x+17=4x+12$

i.e. $3x^2-16x+5=0$

i.e. $3x^2 -15x -x + 5 = 0$

i.e. $3x \left(x-5\right)-1 \left(x-5\right)=0$

i.e. $\left(3x-1\right) \left(x-5\right)=0$

i.e. $x=\dfrac{1}{3}$ or $x=5$

If $x=\dfrac{1}{3}$, then man's present age $=3 \times \left(\dfrac{1}{3}\right)^2 - 12 \times \dfrac{1}{3}+14= 10\dfrac{1}{3}$ years, which is not a reasonable value.

$\therefore$ Daughter's present age $=5$ years

$\therefore$ Man's present age $= 3 \times 5^2 - 12 \times 5 + 14 = 29$ years

Quadratic Equations: Word Problems

A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

Let the unit's digit be $=x$

Since product of the digits $= 18$

$\implies$ $\text{Unit's digit} \times \text{Ten's digit} = 18$

$\implies$ $\text{Ten's digit}= \dfrac{18}{\text{Unit's digit}} = \dfrac{18}{x}$

$\therefore$ The two digit number is $= \dfrac{180}{x}+x$

On subtracting 63 from this number, $\text{Unit's digit} = \dfrac{18}{x}$ and $\text{Ten's digit}=x$

$\therefore$ New two digit number is $= 10x + \dfrac{18}{x}$

$\therefore$ As per question,

$\dfrac{180}{x}+x-63=10x+\dfrac{18}{x}$

i.e. $180+x^2-63x=10x^2+18$

i.e. $9x^2+63x-162=0$

i.e. $x^2+7x-18=0$

i.e. $x^2+9x-2x-18=0$

i.e. $x\left(x+9\right) - 2 \left(x+9\right)=0$

i.e. $\left(x+9\right) \left(x-2\right)=0$

i.e. $x=-9$ or $x=2$

Since unit's digit cannot be negative, $x=-9$ is not a valid answer.

$\therefore$ Unit's digit $=2$

$\therefore$ Ten's digit $= \dfrac{18}{2}=9$

$\therefore$ The number is $9\times10+2 = 92$

Quadratic Equations: Word Problems

A bus covers a distance of 240 km at an uniform speed. Due to heavy rain, its speed gets reduced by 10 kmph and as such it takes two hours longer to cover the distance. Assuming the uniform speed to be x kmph, form an equation and solve it to evaluate x.

Original speed of bus $=x \ kmph$

Distance $=240 \ km$

Time taken to cover 240 km $= \dfrac{240}{x} \ hours$

New speed of bus $=\left(x-10\right) \ kmph$

$\therefore$ New time taken to cover 240 km $= \dfrac{240}{x-10} \ hours$

$\therefore$ As per sum,

$\dfrac{240}{x-10}=\dfrac{240}{x}+2$

i.e. $120x = 120x -1200 + x \left(x-10\right)$

i.e. $x^2 - 10x - 1200 = 0$

i.e. $x^2 - 40x + 30 x -1200=0$

i.e. $x \left(x-40\right) + 30 \left(x-40\right)=0$

i.e. $\left(x-40\right)\left(x+30\right)=0$

i.e. $x=40$ or $x=-30$

Since the speed of the bus cannot be negative

$\therefore$ original speed of the bus $= 40 \ kmph$

Quadratic Equations: Word Problems

The sum of ages of A and B is 47 years. The product of their ages is 550. Find their ages. Assume B to be younger to A.

Let A's age $=x$ years

Since, sum of ages of A and B is 47 years $\implies$ B's age $=\left(47-x\right)$ years

Product of ages of A and B is 550

i.e. $x \left(47-x\right)=550$

i.e. $47x - x^2 = 550$

i.e. $x^2 - 47x + 550 = 0$

i.e. $x^2 - 25x - 22x + 550 = 0$

i.e. $x\left(x-25\right)-22\left(x-25\right)=0$

i.e. $\left(x-25\right)\left(x-22\right)=0$

i.e. $x=25$ or $x=22$

Since B is younger to A,

A's age $=25$ years and B's age $=47-25=22$ years