<s>[INST] Limit of the exponential functions: $\lim_{x\to 0} \frac{e^x-e^{x \cos x}} {x +\sin x}$ I want to find the limit of this function by simply using algebraic manipulation. Though I have computed the limit through L' Hospital's method but still I want to compute the limit purely by function's manipulation to yield a form where limit can be applied 
$$\lim_{x\to 0} \frac{e^x-e^{x \cos x}}  {x +\sin x} $$
Till now we have been taught basic limits such as $\lim_{x\to 0} \frac{e^x-1}{x}=1$ and that's why I have been trying to bring such form in this expression.
P.S. I got the current answer by L'Hospital's rule i.e. $0$
 [/INST] $$\begin{align*}
& \lim_{x \rightarrow 0} \frac{e^x - e^{x\cos x}}{x + \sin x} \\
& = \lim_{x \rightarrow 0} \frac{e^x - 1 + 1 - e^{x\cos x}}{x + \sin x} \\
& = \lim_{x \rightarrow 0} \frac{e^x-1}{x+\sin x}
+ \lim_{x \rightarrow 0} \frac{1-e^{x\cos x}}{x+\sin x} \\
& = \lim_{x \rightarrow 0} \frac{e^x-1}{x} \cdot \frac{x}{x+\sin x}
+ \lim_{x \rightarrow 0} \frac{1-e^{x\cos x}}{-x\cos x} \cdot \frac{-x\cos x}{x+\sin x} \\
& = \lim_{x \rightarrow 0} \frac{e^x-1}{x} \cdot
\lim_{x \rightarrow 0} \frac{1}{1 + \frac{\sin x}{x}}
+ \lim_{x\cos x \rightarrow 0} \frac{e^{x\cos x}-1}{x\cos x} \cdot
\lim_{x \rightarrow 0} -\frac{x}{x+\sin x} \cdot \cos x \\
&  = 1\cdot \frac{1}{1+1} + 1 \cdot -\frac{1}{1+1} \cdot 1\\
&  = \frac{1}{2} - \frac{1}{2}\\
& = 0 \\
\end{align*}$$ 
</s>