I'm trying to find the following limit $$ \lim_{x \to \infty} x(e^{1/x}-1)^x,$$ without L'Hopital or Taylor.

So far I got

\begin{align*} \lim_{x \to \infty} x(e^{1/x}-1)^x &= \lim_{x \to \infty} x(e^{1/x}-1)^x \\ &= \lim_{u \to 0} \frac{1}{\ln(u+1)}(u)^{\frac{1}{\ln(u+1)}} \\ &= \lim_{u \to 0} \frac{1}{\ln(u+1)} \left((u)^{\frac{u}{ \ln(u+1)}}\right)^{\frac{1}{u} } \\ &= \lim_{u \to 0} \frac{1}{\ln(u+1)} \left((u)^{\frac{1}{ \ln(u+1)^{1/u}}}\right)^{\frac{1}{u} } \\ \end{align*}

Any hint for going on?

  • 2
    $\begingroup$ "without L'Hopital or Taylor." What are you allowed to use? For example, what ('standard') limits concerning exponentials and/or logarithms can you rely on; of what definitions? $\endgroup$ – StackTD Nov 9 '18 at 14:21
  • $\begingroup$ $\lim_{x \to \infty}(1+a/x)^x$, and $\lim_{x \to 0}\frac{e^x-1}{x}$, i think.. $\endgroup$ – P. M. O. Nov 9 '18 at 14:28
  • $\begingroup$ @P.M.O. I can't see how it is possible to solve this limits problem without L'Hospital or Taylor if you can only use those two limits... $\endgroup$ – DonAntonio Nov 9 '18 at 14:34
  • $\begingroup$ Maybe other... but without L'h or taylor $\endgroup$ – P. M. O. Nov 9 '18 at 14:37
  • $\begingroup$ Have you tried $y = 1/x$ and $y \to 0$? $\endgroup$ – Stockfish Nov 9 '18 at 14:57

Take logs: $\ln x + x \ln (e^{1/x}-1) \to-\infty$. So limit is $0$.

  • $\begingroup$ The limit of the second addend is imo almost as tricky (hard) as the original one. Again, what the OP can use is important here. $\endgroup$ – DonAntonio Nov 9 '18 at 14:26
  • $\begingroup$ @DonAntonio, $x(\frac{\log(x)}{x} + \log(e^{1/x} -1))$. First summand goes to 0, second goes to $-\infty$ and since x is positive, the whole thing goes to 0. $\endgroup$ – James Yang Nov 9 '18 at 15:02
  • $\begingroup$ @James I'm not sure what first summand you're talking about, but $\;\frac{\log x}x\xrightarrow[x\to0^+]{}-\infty\;$ ... $\endgroup$ – DonAntonio Nov 9 '18 at 15:07
  • $\begingroup$ @DonAntonio, problem states $x \to \infty$. $\endgroup$ – James Yang Nov 9 '18 at 15:11
  • $\begingroup$ @JamesYang You're completely right. Thank you. $\endgroup$ – DonAntonio Nov 9 '18 at 16:02

By standard limit $t^t\to 1$ as $t\to 0$ we have

$$x(e^{1/x}-1)^x=\frac{(e^{1/x}-1)^x}{\frac1x}=\left(\frac{e^{1/x}-1}{\left(\frac1{x}\right)^\frac1x}\right)^x \to \left[\left(\frac{0}{1}\right)^\infty=\right]0$$

  • $\begingroup$ There was no need of $t^t\to 1$ as the numerator clearly tends to $0$ much much faster than denominator. $\endgroup$ – Paramanand Singh Nov 9 '18 at 15:01
  • $\begingroup$ @ParamanandSingh I've pointed it out to make clear the we are not dealing with an indeterminate form $0/0$. I'm not sure the asker has that concept clear. Of course we can also put it in many other different ways. Here we only need to use that standard limit to conlcude. $\endgroup$ – user Nov 9 '18 at 15:03

Your approach is promising and can be easily completed. Just note that $u/\log(1+u)\to 1$ so one can replace first factor $1/\log(1+u)$ with $1/u$ to get $$\lim_{u\to 0^{+}}u^{1/\log(1+u)-1}$$ The exponent tends to $\infty$ and hence the desired limit is $0$. In case you want to be more formal (as some instructors may insist on it) just note that if $0<u<1/3$ we have $\log(1+u)\leq u$ so that $$\frac{1}{\log(1+u)}-1\geq \frac{1}{u}-1\geq 2$$ and therefore expression under limit lies between $0$ and $u^2$ if $0<u<1/3$. Proof is complete now by squeeze theorem.

  • $\begingroup$ That's also a nice approach by another standard limit. $\endgroup$ – user Nov 9 '18 at 15:06
  • $\begingroup$ You're using here, in your first line, one non-trivial limit the OP didn't mention. This is the reason why I said that using only hat her wrote I can't see how this can be done without derivatives in some form...Also the inequalities thereafter are something that usually requires derivatives. $\endgroup$ – DonAntonio Nov 9 '18 at 16:02
  • $\begingroup$ @DonAntonio: don't you think that $u/\log(1+u)\to 1$ is same as $(e^x-1)/x\to 1$? Further the inequalities and squeeze are needed only when someone adamantly demands it. Otherwise it is obvious that $u^{g(u)}\to 0$ if $u\to 0^{+}$ and $g(u) \to\infty$. $\endgroup$ – Paramanand Singh Nov 10 '18 at 4:05
  • $\begingroup$ @ParamanandSingh It doesn't matter what you or I think of whatever. The OP stated the results he can use, that's all I pointed out. And about the inequalitites I meant $$\;\frac1{\log(1+u)}-1\ge\frac1u-1$$ which is probably better proved by means of derivatives, monotonicity and etc. That's why I commented that without Taylor, L'Hospital and etc, and with what the OP said he can use, it's hard for me to see how this problem can be tackled... $\endgroup$ – DonAntonio Nov 10 '18 at 9:12
  • $\begingroup$ @DonAntonio: well those inequalities can be avoided. More simply since $\log(1+u)\to 0$ it follows that $(1/\log(1+u))-1\to\infty $ and hence there is a $k$ with $0<k<1$ such that $(1/\log(1+u))-1>2$ for all $0<u<k$. Thus for $0<u<k$ we have $0<u^{(1/\log(1+u))-1}<u^2$ and squeeze theorem completes the job. $\endgroup$ – Paramanand Singh Nov 11 '18 at 15:04

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