TITLE: Limit of the exponential functions: $\lim_{x\to 0} \frac{e^x-e^{x \cos x}} {x +\sin x}$
QUESTION [6 upvotes]: 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. $0$

REPLY [2 votes]: 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 $x\to 0$.