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$\lim_{x \to 0} \frac{ {(e^x-1)-(e^{\sin x}-1)} } {x-\sin x}$

$ =\lim_{x \to 0} \frac{ \left[ \left( \frac{e^x-1}{x} \right) x \right] -  \left[ \left (\frac{e^{\sin x}-1)}{\sin x} \right) \sin x \right] } {x-\sin x}$

$ =\lim_{x \to 0} \frac{ \left[\ \left(\lim_{x \to 0} \frac{e^x-1}{x} \right) x \right] -  \left[ \left (\lim_{\sin x \to 0}\frac{e^{\sin x}-1)}{\sin x} \right) \sin x \right] } {x-\sin x}$

$=\lim_{x \to 0} \frac{(x-\sin x)}{(x-\sin x)} \\= 1$
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$\hspace{-0.5 cm }$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 $\bf{(x-\sin x)\rightarrow 0}$

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