Answer
\[\lim_{x\rightarrow 0}\left(\frac{x-\sin x}{x-\tan x}\right)=\frac{-1}{2}\]
Work Step by Step
Let \[l=\lim_{x\rightarrow 0}\left(\frac{x-\sin x}{x-\tan x}\right)\]
Which is $\frac{0}{0}$ form
Using L'Hopital's rule
\[l=\lim_{x\rightarrow 0}\frac{(x-\sin x)'}{(x-\tan x)'}\]
\[l=\lim_{x\rightarrow 0}\frac{(1-\cos x)}{(1-\sec^2 x)}\]
Which is again $\frac{0}{0}$ form
Using L' Hopital's rule
\[l=\lim_{x\rightarrow 0}\frac{(1-\cos x)'}{(1-\sec^2 x)'}\]
\[\Rightarrow l=\lim_{x\rightarrow 0}\frac{\sin x}{(-2\sec^2 x\tan x)}\]
\[\Rightarrow l=\lim_{x\rightarrow 0}\frac{\sin x\cos x}{(-2\sec^2 x\sin x)}\]
\[\Rightarrow l=\lim_{x\rightarrow 0}\frac{\cos x}{(-2\sec^2 x)}\]
\[\Rightarrow l=\frac{-1}{2}\]
Hence , \[\lim_{x\rightarrow 0}\left(\frac{x-\sin x}{x-\tan x}\right)=\frac{-1}{2}\]
