Evaluate $\lim\limits_{x \to 0} \; \dfrac{x^2 e^x}{\tan^2 x }$
$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{x^2 e^x}{\tan^2 x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(x^2 e^x\right)}{\dfrac{d}{dx}\tan^2 x} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{x^2 e^x + 2x e^x}{2 \tan x \sec^2 x} & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{xe^x \left(x+2\right)}{2 \tan x \sec^2 x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left[xe^x \left(x+2\right)\right]}{\dfrac{d}{dx}\left(2 \tan x \sec^2 x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{xe^x + \left(x+2\right)\left(xe^x + e^x\right)}{2 \left(\tan x \times 2\sec x \times \sec x \tan x + \sec^2 x \times \sec^2 x\right) } & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{4xe^x + x^2 e^x + 2e^x}{4 \tan^2 x \sec^2 x + 2 \sec^4 x} & \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{e^x\left(x^2+4x+2\right)}{4\sec^2 x \left(\sec^2 x -1\right)+2\sec^4 x} & \\ & & \\ & = \dfrac{1\left(0+0+2\right)}{4 \times 1 \times \left(1-1\right)+ 2 \times 1} = \dfrac{2}{2} = 1 \end{aligned}$