Mathematics - Limits Question with Solution | TestHub

LHL $={\lim }_{x\to 0^{-}}f(0-h)={\lim }_{h\to 0}\frac{\sin [\cos h]}{1+[\cos h]}=\frac{\sin (0)}{1+0}=0\left(\genfrac{}{}{0ex}{}{\because \mathrm{h}>0}{\therefore \cos \mathrm{h}<1}\right)$

RHL $={\lim }_{x\to 0^{+}}f(x)={\lim }_{h\to 0}f(0+h)={\lim }_{h\to 0}\frac{\sin [\cos h]}{1+[\cos h]}=\frac{\sin (0)}{1+0}=0\left(\genfrac{}{}{0ex}{}{\because \mathrm{h}>0}{\therefore \cos \mathrm{h}<1}\right)$

$\therefore {\lim }_{x\to 0}\frac{\sin [\cos x]}{1+[\cos x]}=0$