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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!

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  • $\begingroup$ Welcome to MSE. You should always type all the relevant information to the question yourself (here is how to format maths guide), since external links can become broken, and this question won't be searchable, thus won't help anyone in the future. $\endgroup$ – Ennar Jan 3 '17 at 16:43
  • $\begingroup$ On another note, what is sen x? $\endgroup$ – Ennar Jan 3 '17 at 16:44
  • $\begingroup$ Is that $$\lim_{x\to0}\frac{\log\frac{e^{x^2}-1}{x^2}}{x\sin x}$$ $\endgroup$ – Nosrati Jan 3 '17 at 16:44
  • $\begingroup$ I already included the content of the link to the answer. I guess Manuel is spaniard like me, we use $sen x$ here instead of $sin$ when working in our native language. $\endgroup$ – kub0x Jan 3 '17 at 16:46
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    $\begingroup$ Applying L'Hospital twice yields the wanted result. Not really nice since the numerator gets quite long but it works. $\endgroup$ – Fritz Jan 3 '17 at 16:48
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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 $x^2 \rightarrow t$, $$ \begin{align} \lim_{t\to 0} \frac{\log \left(\frac{e ^ {t} - 1}{t} \right)}{t} &= \lim_{t\to 0} \frac{\log \left(1 + \frac{e ^ {t} - 1 - t}{t} \right)}{\frac{e ^ {t} - 1 - t}{t}} \times \frac{e ^ {t} - 1 - t}{t^2} \\ &= 1 \times \frac{1}{2} = \frac{1}{2} \end{align}$$

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  • $\begingroup$ Thanks, Guru! I hadn't thought about that transformation, but it makes sense and yields the same result as Wolfram Alpha. $\endgroup$ – Manuel Jan 3 '17 at 16:58
  • $\begingroup$ You're welcome, don't forget to mark the answer to your question as accepted. $\endgroup$ – Blencer Jan 3 '17 at 16:59
  • $\begingroup$ I don't think I completely understand something. Why do you add and substract 1? It's the last step that I don't understand. How do you infer that all those things equal 1 times 1/2? Edit: okay, I get it now haha. Thanks! $\endgroup$ – Manuel Jan 4 '17 at 18:15
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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.

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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

$$\lim_{x \to 0} \frac{\log\left(\frac{e^{x^2}-1}{x^2}\right)}{x\sin(x)} = \frac12.$$

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Hint

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

  • put $x^2=t$

  • use $\frac{\ln(1+X)}{X}\sim X \;(X\to 0)$.

with $X=\frac{e^x-1}{x}-1$.

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  • $\begingroup$ Is $\lim\frac{\ln(1+x)}{x}\to1$ equivalent to $\lim x\to0$ as $x\to0$? $\endgroup$ – Nosrati Jan 3 '17 at 16:54

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