Find $\displaystyle\lim_{n\to\infty} n(e^{\frac 1 n}-1)$

This should be solved without LHR. I tried to substitute $n=1/k$ but still get indeterminant form like $\displaystyle\lim_{k\to 0} \frac {e^k-1} k$. Is there a way to solve it without LHR nor Taylor or integrals ?

Maybe with the definition of a limit ?


$f(x)'=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{(x+h)(e^{1/x+h}-1)-x(e^{\frac 1 x}-1)}{h}= \lim_{h\to 0}\frac{xe^{1/x+h}+he^{1/x+h}-x-h-xe^{\frac 1 x}+x}{h}= \lim_{h\to 0}\frac{xe^{1/x+h}+he^{1/x+h}-h-xe^{\frac 1 x}}{h}$

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    $\begingroup$ It should be $\frac{e^k-1}{k}$. Which definition of $e^x$ are you working with? $\endgroup$ – Daniel Fischer May 20 '14 at 13:39
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    $\begingroup$ Write it as ${e^{1/n}-e^0\over 1/n}$ and try to recognize the limit as a derivative. $\endgroup$ – David Mitra May 20 '14 at 13:41
  • $\begingroup$ @DanielFischer right, fixed. I think we learned all the definitions of it. $\endgroup$ – GinKin May 20 '14 at 13:43
  • $\begingroup$ So write $$e^x = \sum_{m=0}^\infty \frac{x^m}{m!},$$ compute $n(e^{1/n}-1)$ from it, and take the limit. $\endgroup$ – Daniel Fischer May 20 '14 at 13:45
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    $\begingroup$ @DanielFischer but that's Taylor and I'm not supposed to use it here... $\endgroup$ – GinKin May 20 '14 at 13:46

Why should one use Taylor where we don't need it at all?

$$L = \lim\limits_{n\to\infty}n\left(e^\frac{1}{n}-1\right)=\lim\limits_{x\to0}\frac{e^x-1}{x}$$

Substitute $u=e^x-1$. Then $x=\ln(u+1)$

$$L=\lim\limits_{u \to 0}\frac{u}{\ln(1+u)}=\lim\limits_{u\to0}\frac{1}{\frac{1}{u}\ln(1+u)}=\lim\limits_{u\to0} \frac{1}{\ln \left ( 1+u\right)^{1/u}}=\frac{\lim\limits_{u\to0} 1}{\lim\limits_{v \to \infty} \ln\left(1+\frac{1}{v}\right)^v}=\frac{1}{\ln e}=1$$

  • $\begingroup$ Very neat! Nice job. $\endgroup$ – Deepak May 20 '14 at 14:33

Hint: You have $\lim_{n\to\infty}\dfrac{e^\frac{1}{n}-e^0}{\frac{1}{n}}=\lim_{k\to0}\dfrac{e^k-e^0}{k}$. Use the definition of a differential.

  • $\begingroup$ You mean $f(x)'=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ ? I tried it now and it didn't work... $\endgroup$ – GinKin May 20 '14 at 13:54
  • $\begingroup$ @GinKin What did you try? $\endgroup$ – user122283 May 20 '14 at 13:56
  • $\begingroup$ Please see the edit. $\endgroup$ – GinKin May 20 '14 at 14:04
  • $\begingroup$ @GinKin Interpret as follows: $$e^k=e^{k+0}$$ You'll get it now. $\endgroup$ – user122283 May 20 '14 at 14:06
  • $\begingroup$ But if I substitute h with 0 I still get 0/0... $\endgroup$ – GinKin May 20 '14 at 14:14

Put $x = \frac{1}{n}$

The limit becomes $\lim_{x \to 0}\frac{e^x-1}{x} \rightarrow \lim_{x \to 0}\frac{1+x-1}{x} = 1$

using the Maclaurin series for $e^x$

EDIT: Just noticed you also excluded Taylor series (which would preclude Maclaurin's) in your question. So feel free to ignore my answer.

  • $\begingroup$ No Taylor.${}{}$ $\endgroup$ – user122283 May 20 '14 at 13:41
  • $\begingroup$ Yeah, just noticed that. Jumped right in after the first line which said "without LHR", so I assumed Maclaurin's/Taylor's was OK. Then I read the rest. :) $\endgroup$ – Deepak May 20 '14 at 13:44
  • $\begingroup$ It is OK since even without Taylor one can prove that $e^x - 1\sim x$ as $x\to 0$ $\endgroup$ – Тимофей Ломоносов May 20 '14 at 14:14

Again this turns out to be a very nice question. Without making any assumptions on $e$ or $e^{x}$ it is possible to show that the limit $\lim_{n \to \infty}n(e^{1/n} - 1)$ exists. To be more general we can show that for any real number $x > 0$ the limit $$f(x) = \lim_{n \to \infty}n(x^{1/n} - 1)$$ exists. We need to deal with cases $0 < x < 1$, $x = 1$ and $x > 1$ and to it can be seen that the case $0 < x < 1$ can be handled via the case $x > 1$ if we put $x = 1/y$. For $x = 1$ the limit is obviously $0$.

For $x > 1$ we need to show that the sequence $g(x, n) = n(x^{1/n} - 1)$ decreases as $n$ increases and obvious $g(x, n) > 0$ so that the limit $f(x) = \lim_{n \to \infty}n(x^{1/n} - 1)$ exists.

As a next step we can show that the limit function $f(x)$ is a strictly increasing function of $x$ for $x > 0$ and satisfies $$f(1) = 0, f(xy) = f(x) + f(y)$$ and with some more effort we can show that $f(x)$ is differentiable and $f'(x) = 1/x$ so that $f(x)$ has all the properties of $\log x$ or $\ln x$.

If we define $e$ as a number such that $\log e = 1$ then obviously we get $\lim_{n \to \infty}n(e^{1/n} - 1) = 1$. This approach towards the definition of logarithm via limits is presented in detail in my blog post.

  • $\begingroup$ Do you use multi variable calculus here ? $\endgroup$ – GinKin May 26 '14 at 10:47
  • $\begingroup$ No its all single variable calculus. see my blog post linked in the answer. $\endgroup$ – Paramanand Singh May 26 '14 at 10:57

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