Question

Use the sinx}{x}=1" alt="lim_{x-0} \frac{sinx}{x}=1" align="absmiddle" class="latex-formula"> to determine $lim_{x-0} \frac{xcos5x}{sin5x}$.

Answer 1

Rewrite the limit as

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

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

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