5

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

Winther
  • 24,478
user12
  • 705
  • 2
    Hint: The stuff in the parentheses tends to $2$ – Alex G. Dec 29 '15 at 18:17
  • 1
    Use $h=1/x^{2}$. Thus $h\rightarrow +\infty$ as $x\rightarrow 0$. – Albert Dec 29 '15 at 18:17
  • why the parentheses tends to 2? – user12 Dec 29 '15 at 18:23
  • You can plug this into wolfram by the way. The answer is that there is no limit. The question is how to get there. – pancini Dec 29 '15 at 18:31
  • You are asking another question right now. Please open another thread about it, and be careful with parenthesis next time :) – Paolo Leonetti Dec 29 '15 at 18:53
  • 1
    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 ${e^{2x}-1\over x^2}$. – ForgotALot Dec 29 '15 at 19:33
  • Several of the answers have interpreted the question as it was originally written: $e^{2x}$ instead of $e^{x^2}$. By changing the question now (even if this is what you intended to ask) you invalidate the answers given. I think you should just keep it as it was with $e^{2x}$ and if you can't see how to solve the $e^{x^2}$ problem from these answers you can ask a new question. – Winther Jan 06 '16 at 07:39
  • @Winther I'm sorry, that was my mistake - the question popped up in a review queue, and I edited without looking at the fact that the answers had already interpreted it a different way. That's why I rolled back my own edit. –  Jan 06 '16 at 07:39
  • To solve your new question notice that $e^{x^2} \approx 1 + x^2 + \frac{x^4}{2} + O(x^6)$ so $\left(\frac{e^{x^2}-1}{x^2}\right)^{\frac{1}{x^2}} \sim \left((1+x^2/2)^{\frac{2}{x^2}}\right)^{1/2}$. You can now try to use $e = \lim_{x\to 0}(1+x)^{1/x}$ to evaluate the limit. – Winther Jan 06 '16 at 07:44

2 Answers2

3

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.

Paolo Leonetti
  • 15,423
  • 3
  • 24
  • 57
  • 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?) – Sasha Dec 29 '15 at 18:46
  • 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.. – Paolo Leonetti Dec 29 '15 at 18:52
  • 1
    @jordan i deleted the comment, looks good now – hunter Dec 29 '15 at 18:55
  • 1
    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. – Sasha Dec 29 '15 at 18:57
0

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

GaussTheBauss
  • 2,283