Evaluate the following limit (1)

We have an indeterminate $\dfrac{0}{0}$ situation, so applying L'Hôpital's rule,

$\begin{aligned}\lim_{x\to 0}\frac{4x}{\tan x}&=\lim_{x\to 0} \frac{4}{\sec^2 x}\\&=4\cos^2 0\\&=\color{#20A900}{\boxed{4}}\end{aligned}$

$\lim\limits_{x\to 0}\frac{4x}{\tan(x)} = \lim\limits_{x\to 0}\frac{\frac{d}{dx}(4x)}{\frac{d}{dx}(\tan(x))}\: \left( \textrm{L'Hopital's rule: }\frac{0}{0}\right) = \lim\limits_{x\to 0}\frac{4}{\sec^2(x)}$ $\lim\limits_{x\to 0}\frac{4}{\sec^2(x)} = \frac{4}{1^2} = \boxed{4}$