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