\(\displaystyle\lim_{x\to0}\frac{\sin5x}{\tan2x}=\lim_{x\to0}\frac{5x}{5x}\frac{2x}{2x}\frac{\sin5x\cos2x}{\sin2x}\)
\(\displaystyle=\left(\lim_{x\to0}\frac{\sin5x}{5x}\right)\left(\lim_{x\to0}\frac{2x}{\sin2x}\right)\left(\lim_{x\to0}\frac{2x\cos2x}{5x}\right)\)
The first two limits involve a property you should be familiar with; they both evaluate to 1. In the last limit, the \(x\)'s cancel and since \(\cos2x\) is continuous, you're left with
\(\displaystyle\lim_{x\to0}\dfrac2\cos2x}5=\frac25\cos\left(\lim_{x\to0}2x\right)=\dfrac25\cos0=\dfrac25\)
