I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Can you explain the method and the steps used? Thanks

$$\lim\limits_{x \to 0} \left(\frac{e^{x^2} -1}{x^2}\right)^\frac{1}{x^2}$$

Note: In a previous version of this question the limit was written as $\left(\frac{(e^{x})^2 -1}{x^2}\right)^\frac{1}{x^2}$.

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    $\begingroup$ Hint: The stuff in the parentheses tends to $2$ $\endgroup$ – Alex G. Dec 29 '15 at 18:17
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    $\begingroup$ Use $h=1/x^{2}$. Thus $h\rightarrow +\infty$ as $x\rightarrow 0$. $\endgroup$ – Albert Dec 29 '15 at 18:17
  • $\begingroup$ why the parentheses tends to 2? $\endgroup$ – user12 Dec 29 '15 at 18:23
  • $\begingroup$ You can plug this into wolfram by the way. The answer is that there is no limit. The question is how to get there. $\endgroup$ – Elliot G Dec 29 '15 at 18:31
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    $\begingroup$ Would it be possible for the original author Amarildo to write the expression in parentheses correctly in MathJax, perhaps in a comment? I read it as ${e^{x^2}-1\over x^2}$, but I see an answer that seems to take it to be ${e^{2x}-1\over x^2}$. $\endgroup$ – ForgotALot Dec 29 '15 at 19:33

Right limit. Set $y=x^{-1}$ then as $y\to \infty$ we obtain $$ \left(y^2 \left(e^{2/y}-1\right)\right)^{y^2}\approx \left(y^2\cdot \frac{2}{y}\right)^{y^2} \to \infty. $$

Left limit. Suppose now that $y=-x^{-1}$. Then as $y\to \infty$ it holds $$ \left(y^2 \left(e^{-2/y}-1\right)\right)^{-y^2}\approx \left(y^2\cdot \frac{-2}{y}\right)^{-y^2}=\left(\frac{1}{2y}\right)^{y^2} \to 0. $$

The limits are different, therefore it does not exist.

  • $\begingroup$ Thanks. Despite original mistake, your answer is much shorter, and thus easier to read than others. Still I want to ask (I'm not mathematician, so I don't argue, I just ask): (1) Isn't it a mistake to write ≈ instead of ~? (2) Isn't it a mistake to write y→∞ instead of y→+∞ or y→∞⁺? (Isn't y→∞ supposed to mean two-sided limit?) $\endgroup$ – Sasha Dec 29 '15 at 18:46
  • $\begingroup$ Thanks for the comment :) With $y\to \infty$ I always mean here $y\to +\infty$, but it is just a matter of convention. About your other notational question, yes, it is correct: $f(x)\sim g(x)$ is equivalent to $f(x)=g(x)(1+o(1))$ as $x\to +\infty$, but I just wanted to avoid small o notation and keep it as intuitive as possible.. $\endgroup$ – Paolo Leonetti Dec 29 '15 at 18:52
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    $\begingroup$ @jordan i deleted the comment, looks good now $\endgroup$ – hunter Dec 29 '15 at 18:55
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    $\begingroup$ Aha, thanks. I just noticed that it contradicts conventions of our school (we always used y→∞ to mean two-side and not used ≈ in this case), so I just was interested on how our school conventions relate to international ones. Thank you. $\endgroup$ – Sasha Dec 29 '15 at 18:57


$$y = \left(\frac{e^{2x} -1}{x^2}\right)^\frac{1}{x^2}.$$


$$\ln y = \frac{1}{x^2} \ln\left(\frac{e^{2x} -1}{x^2}\right).$$

Notice that

$$\lim_ \limits{x \to 0^+} \ln\left(\frac{e^{2x} -1}{x^2}\right) = \lim_ \limits{x \to 0^+} \ln\left(\frac{1+2x+4x^2+\cdots-1}{x^2}\right)=\infty,$$

but $$\lim_ \limits{x \to 0^-} \ln\left(\frac{e^{2x} -1}{x^2}\right) = \lim_ \limits{x \to 0^-} \ln\left(\frac{1+2x+4x^2+\cdots-1}{x^2}\right)$$

is undefined since you have $-\infty$ inside the $\log$.

Therefore the limit does not exist


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