Question

Use the
`lim_(x-0) (sinx)/(x)=1` to determine
`lim_(x-0) (xcos5x)/(sin5x)`.

Answer 1

Rewrite the limit as

`\displaystyle \lim_(x\to0) (x\cos(5x))/(\sin(5x)) = \lim_(x\to0) (5x)/(\sin(5x)) \cdot \lim_(x\to0) \frac{\cos(5x)}5`

Then using the known limit,

`\displaystyle \lim_(x\to0) \frac{\sin(x)}x = 1 \implies \frac1{\lim\limits_(x\to0)\frac{\sin(x)}x} = \lim_(x\to0)\frac x{\sin(x)}=1`

it follows that

`\displaystyle \lim_(x\to0) (x\cos(5x))/(\sin(5x)) = 1 \cdot \frac{\cos(0)}5 = \boxed{\frac15}`