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(C) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}$ which is of the form $\frac{0}{0}$.Using $L'Hospital's

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(C) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}$ which is of the form $\frac{0}{0}$.Using $L'Hospital's$ rule:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}({e^x} - {e^{\sin x}})}{\frac{d}{dx}(x - \sin x)} = \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{\sin x}}\cos x}}{{1 - \cos x}}$Applying $L'Hospital's$ rule again:$\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - ({e^{\sin x}}(-\sin x) + \cos x({e^{\sin x}}\cos x))}}{{\sin x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} + {e^{\sin x}}\sin x - {e^{\sin x}}\cos^2 x}}{{\sin x}}$Since $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$ and $\cos x 	o 1$,we can simplify the expression using Taylor series expansion for $e^x \approx 1+x+\frac{x^2}{2}$ and $\sin x \approx x - \frac{x^3}{6}$:$\mathop {\lim }\limits_{x 	o 0} \frac{(1+x+\frac{x^2}{2}) - (1+(x-\frac{x^3}{6})+\frac{(x-\frac{x^3}{6})^2}{2})}{x-(x-\frac{x^3}{6})} = \mathop {\lim }\limits_{x 	o 0} \frac{1+x+\frac{x^2}{2} - 1 - x + \frac{x^3}{6} - \frac{x^2}{2}}{\frac{x^3}{6}} = \mathop {\lim }\limits_{x 	o 0} \frac{\frac{x^3}{6}}{\frac{x^3}{6}} = 1$.

$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}} \right]$ is equal to

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