$\hspace{-0.5 cm }$Given $\displaystyle \bf{\lim_{x\rightarrow 0}\frac{e^x-\sin x}{x-\sin x} = \lim_{x\rightarrow 0}\left[e^{\sin x}\cdot \frac{e^{x-\sin x}-1}{x-\sin x}\right]}$
So $\displaystyle \bf{\lim_{x\rightarrow 0}e^{\sin x}\times \lim_{x\rightarrow 0}\frac{e^{x-\sin x}-1}{x-\sin x} = 1\times 1=1}$
Above we have used $\displaystyle \bf{\lim_{y\rightarrow 0}\frac{e^y-1}{y} = 1}$, Bcz above when $\bf{x\rightarrow 0}$
Then $\bf{(x-\sin x)\rightarrow 0}$