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