5
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Hint: The stuff in the parentheses tends to $2$ $\endgroup$ – Alex G. Dec 29 '15 at 18:17
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @jordan i deleted the comment, looks good now $\endgroup$ – hunter Dec 29 '15 at 18:55
  • 1
    $\begingroup$ 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
0
$\begingroup$

Let

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

Then

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.