To prove $\lim_{x\to 0}\frac{e^x-e^{x\cos x}}{x-\sin x}=3$ I came across this question to prove the given limit
$$\lim_{x\to 0}\frac{e^x-e^{x\cos x}}{x-\sin x}=3$$
First I tried using LHospital's rule directly.
Then I tried using expansion of $e^x$ and then using LHospital's rule but I am getting stuck.

Use $\cos(x) = 1 - \frac12 x^2 + \cal{O}(x^4)$ and $\sin(x) = x - \frac{x^3}{6} + \cal{O}(x^5)$. Then 
$$
\lim_{x\to 0}\frac{e^x-e^{x\cos x}}{x-\sin x}=\lim_{x\to 0}e^x\frac{1-e^{-x^3/2 + \cal{O}(x^5)}}{x-\sin x}= \lim_{x\to 0}\frac{{x^3/2  + \cal{O}(x^5)}}{x^3/6 + \cal{O}(x^5)}=\lim_{x\to 0}\frac{{1/2  + \cal{O}(x^2)}}{1/6 + \cal{O}(x^2)}= 3
$$
