Q:

How to understand the following limit?

I am an beginner and calculating the following limit:
$$\lim_{x\rightarrow0} \left(\frac{e^x-e^{\sin x}}{x-\sin x}\right)$$
The correct answer is $1$ but I cannot understand it:
If $x$ is close to $0$, then $e^x = 1$, $e^{\sin x} = 1$, $x = 0$ and $\sin x = 0$.
Aren't we dealing with $\left(\frac{0}{0}\right)$? How to get $1$?
Any help would be appreciated.

Thanks for your replies, now I can solve it!
Indeed, when x->0, we are dealing with $\left(\frac{0}{0}\right)$, so we need to apply L' Hopital's Rule: assume $t = x - \sin x$,
$$e^{\sin x} \frac{(e^t - e^0)'}{(t - 0)'}  = e^{\sin x}\frac{e^t - 0}{1 - 0} $$
when x->0 then t->0 so $$e^{\sin x}\frac{e^t - 0}{1 - 0} = e^{sin0}\frac{1-0}{1} = 1$$

A:

You can factor $e^{\sin x}$ out of the numerator to get
$$ e^{\sin x} \left( \frac{e^{x - \sin x} - 1}{x - \sin x} \right)$$ and write $t = x - \sin x$ to further get $$e^{\sin x} \left( \frac{e^t - e^0}{t - 0} \right).$$
As $x \to 0$ you also have $t \to 0$ so by continuity and some limit laws the limit equals $$e^{\sin 0} \cdot \left. \frac{d}{dt} e^t  \right|_{t=0} = e^{\sin 0} \cdot e^0 = 1.$$

