Limit as x goes towards 0 for the function (e^x-1)^2/(sin(e^x-1))
Any help would be much appreciated
To simplify things let
$\displaystyle u=e^x-1$
If we take the limit as $\displaystyle x \to 0$ this is the same as the limit as $\displaystyle u \to 0$
Now this gives you the limit
$\displaystyle \lim_{x \to 0}\frac{(e^{x}-1)^2}{\sin(e^x-1)}=\lim_{u \to 0}\frac{u^2}{\sin(u)}$
This gives
$\displaystyle \lim_{u \to 0}\frac{u}{\sin(u)}\cdot u=1 \cdot 0 =0$