${\lim }_{x\to 0}\frac{e^x-e^{\sin x}}{x-\sin x}$

Solution

1. Recall the Taylor series expansions: $e^u=1+u+\frac{u^2}{2}+O(u^3)$ and $\sin x=x-\frac{x^3}{6}+O(x^5)$.
2. Substitute these into the numerator: $$\begin{array}{rl}{e}^x-e^{\sin x}\approx (1+x+\frac{x^2}{2})-(1+(x-\frac{x^3}{6})+\frac{(x-\frac{x^3}{6}{)}^2}{2})& =(1+x+\frac{x^2}{2})-(1+x-\frac{x^3}{6}+\frac{x^2}{2})\\ & =\frac{x^3}{6}+O(x^4)\end{array}$$.
3. Substitute the denominator: $x-\sin x\approx x-(x-\frac{x^3}{6})=\frac{x^3}{6}$.
4. Evaluate the limit: ${\lim }_{x\to 0}\frac{x^3/6}{x^3/6}=1$.
5. Hence the answer is (B).