Limit of the exponential functions: $\lim_{x\to 0} \frac{e^x-e^{x \cos x}} {x +\sin x}$

$$\begin{align*} & \lim_{x \rightarrow 0} \frac{e^x - e^{x\cos x}}{x + \sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x - 1 + 1 - e^{x\cos x}}{x + \sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x-1}{x+\sin x} + \lim_{x \rightarrow 0} \frac{1-e^{x\cos x}}{x+\sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x-1}{x} \cdot \frac{x}{x+\sin x} + \lim_{x \rightarrow 0} \frac{1-e^{x\cos x}}{-x\cos x} \cdot \frac{-x\cos x}{x+\sin x} \\ & = \lim_{x \rightarrow 0} \frac{e^x-1}{x} \cdot \lim_{x \rightarrow 0} \frac{1}{1 + \frac{\sin x}{x}} + \lim_{x\cos x \rightarrow 0} \frac{e^{x\cos x}-1}{x\cos x} \cdot \lim_{x \rightarrow 0} -\frac{x}{x+\sin x} \cdot \cos x \\ & = 1\cdot \frac{1}{1+1} + 1 \cdot -\frac{1}{1+1} \cdot 1\\ & = \frac{1}{2} - \frac{1}{2}\\ & = 0 \\ \end{align*}$$

$$\frac{e^x-e^{x \cos x}} {x +\sin x}=\frac{\frac{e^x}{x}-\frac{e^{x \cos x}}{x}}{1+\frac{\sin x}{x}}=\frac{\frac{e^x-1}{x}-\frac{e^{x \cos x}-1}{x}}{1+\frac{\sin x}{x}}=\frac{\frac{e^x-1}{x}-\cos x\frac{e^{x \cos x}-1}{x \cos x}}{1+\frac{\sin x}{x}}$$

Hence the limit is $\frac{1-1(1)}{1+1}=0$.

Note that $$\frac{e^{x} - e^{x\cos x}} {x+\sin x} = e^{x\cos x} \cdot\frac{e^{x(1-\cos x)} - 1}{x(1-\cos x)} \cdot\dfrac{1-\cos x} {1 + \dfrac{\sin x} {x}} $$ which tends to $$e^{0\cdot 1}\cdot 1\cdot\frac{1-1}{1+1}=0$$ as $x\to 0$.