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