Question

Prove that:
$\underset{x\to 0}{lim}\frac{ax+x\cos x}{b\sin x}$

Answer 1

$={lim}_{x\to 0}\frac{ax+xcosx}{bsinx}$

$={lim}_{x\to 0}\frac{x(a+cosx)}{bsinx}$

$={lim}_{x\to 0}\frac{a+cosx}{b}\times \frac{x}{sinx}$

$={lim}_{x\to 0}\left(\frac{a+cosx}{b}\right)÷{lim}_{x\to 0}\frac{sinx}{x}$

$={lim}_{x\to 0}\frac{a+cosx}{b}÷1$

$={lim}_{x\to 0}\frac{a+cosx}{b}$

put ${lim}_{x\to 0}$

$=\frac{a+cos0}{b}$

$=\frac{a+1}{b}$