Question

${lim}_{x\to 0}\frac{2x-sinx}{tanx+x}$

Answer 1

${lim}_{x\to 0}\frac{2x-sinx}{tanx+x}$

Dividing numerator and denominator by x

$={lim}_{x\to 0}\frac{2-\frac{sinx}{x}}{\frac{tanx}{x}+1}$

$=\frac{{lim}_{x\to 0}2-{lim}_{x\to 0}\frac{sinx}{x}}{{lim}_{x\to 0}\frac{tanx}{x}+{lim}_{x\to 0}1}$

$=\frac{2-1}{1+1}[\because {lim}_{x\to 0}\frac{sinx}{x}=1,{lim}_{x\to 0}\frac{tanx}{x}=1]$

$=\frac{1}{2}$