Question

Lim x->0 sin5x/tan2x

Answer 1

`\displaystyle\lim_(x\to0)(\sin5x)/(\tan2x)=\lim_(x\to0)(5x)/(5x)(2x)/(2x)(\sin5x\cos2x)/(\sin2x)`

`\displaystyle=\left(\lim_(x\to0)(\sin5x)/(5x)\right)\left(\lim_(x\to0)(2x)/(\sin2x)\right)\left(\lim_(x\to0)(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)\frac2\cos2x}5=\frac25\cos\left(\lim_(x\to0)2x\right)=\frac25\cos0=\frac25`