If $\lim\limits_{x \to 0} \; \dfrac{\sin 2x + k \sin x}{x^3}$ is finite, find k and the limit.
$\begin{aligned}
& \lim\limits_{x \to 0} \; \dfrac{\sin 2x + k \sin x}{x^3} & \left(\dfrac{0}{0} \text{ form}\right) \\
& & \\
& = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\sin 2x + k \sin x\right)}{\dfrac{d}{dx}\left(x^3\right)} & \left[\text{L'Hospital's rule}\right] \\
& & \\
& = \lim\limits_{x \to 0} \; \dfrac{2 \cos 2x + k \cos x}{3x^2} & \cdots (1)
\end{aligned}$
Equation (1) will be of $\dfrac{0}{0}$ form if $\;\;$ $2 \cos 0 + \cos 0 = 0$
i.e. $2+k = 0$ $\implies$ $k = -2$
$\therefore$ $\;$ Equation (1) becomes
$\begin{aligned}
& \lim\limits_{x \to 0} \; \dfrac{2 \cos 2x - 2 \cos x}{3x^2} & \left(\dfrac{0}{0} \text{ form}\right) \\
& & \\
& = \dfrac{2}{3} \; \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\cos 2x - \cos x\right)}{\dfrac{d}{dx}\left(x^2\right)} & \left[\text{L'Hospital's rule}\right] \\
& & \\
& = \dfrac{2}{3} \; \lim\limits_{x \to 0} \; \dfrac{-2\sin 2x + \sin x}{2x} & \left(\dfrac{0}{0} \text{ form}\right) \\
& & \\
& = \dfrac{2}{3} \; \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(-2\sin 2x + \sin x\right)}{\dfrac{d}{dx}\left(2x\right)} & \left[\text{L'Hospital's rule}\right] \\
& & \\
& = \dfrac{2}{3} \; \lim\limits_{x \to 0} \; \dfrac{-4\cos 2x + \cos x}{2} & \\
& & \\
& = \dfrac{1}{3} \times \left(-4\cos 0 + \cos 0\right) & \\
& & \\
& = \dfrac{1}{3} \times \left(-4+1\right) = -1
\end{aligned}$