Solving simple limit. I've been trying to solve the following limit using different approaches (L'Hôpital, asymptotic equivalences) but I can't get to the right answer. Wolfram Alpha returns $\frac{1}{2}$ as the answer, and so does my calculator when I insert small values for $x$. The step-by-step solution isn't available though.
$$\lim_{x\to 0}\frac{\log(\frac{e^{x^2}-1}{x^2})}{x\sin x}$$
Thanks in advance!
 A: You have :
$$ \lim_{x \to 0} \frac{\log \left(\frac{e ^ {x^2} - 1}{x^2} \right)}{x\sin x} 
= \lim_{x \to 0} \frac{\log \left(\frac{e ^ {x^2} - 1}{x^2} \right)}{x^2} \times \frac{x}{\sin x}
$$
We know that $\displaystyle{\lim_{x \to 0} \frac{x}{\sin x} = 1}$. Now, substituting $x^2 \rightarrow t$, $$ \begin{align} \lim_{t\to 0} \frac{\log \left(\frac{e ^ {t} - 1}{t} \right)}{t} &= \lim_{t\to 0} \frac{\log \left(1 + \frac{e ^ {t} - 1 - t}{t} \right)}{\frac{e ^ {t} - 1 - t}{t}} \times \frac{e ^ {t} - 1 - t}{t^2} \\ &= 1 \times \frac{1}{2} = \frac{1}{2} \end{align}$$ 
A: Hint


*

*replace $\sin(x)$ by its equivalent $x$.

*put $x^2=t$

*use $\frac{\ln(1+X)}{X}\sim X \;(X\to 0)$.
with $X=\frac{e^x-1}{x}-1$.
A: Explaining things a little bit, if $x \to 0 \ e^{x^2} \sim 1+x^2 + \frac{x^4}{2} + O(x^8)$. IF you make this expansion, the rest comes out easily once you use $\log(1+t) \sim t \to \frac{\log (1+t)}{t} \to _t 1$.
This is the result of Maclaurin series expansion.
A: I am converting my comment into an answer. You can actually use L'Hospital to evaluate your limit. But you have to apply it twice. This can be seen as differentiating the denominator twice yields $$(x\sin(x))''=2\cos(x)-x\sin(x)$$ which converges to $2$ for $x \to 0$.
Applying L'Hospital the first time is legit since $x\sin(x) \to 0$ and $\log\left(\frac{e^{x^2}-1}{x^2}\right) \to 0$ for $x\to 0$ as 
$$\lim_{x\to 0} \frac{e^{x^2}-1}{x^2}=\lim_{x \to 0} \frac{2x e^{x^2}}{2x}=1.$$
The same can be seen when applying L'Hospital for the second time. I leave that to you. Finally you get
$$\lim_{x \to 0} \frac{\log\left(\frac{e^{x^2}-1}{x^2}\right)}{x\sin(x)} = \frac12.$$
