Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

show that if $x $ is an element of $\mathbb R$ then $$\lim_{n\to\infty} \left(1 + \frac xn\right)^n = e^x $$

(HINT: Take logs and use L'Hospital's Rule)

i'm not too sure how to go about answer this or putting it in the form $\frac{f'(x)}{g'(x)}$ in order to apply L'Hospitals Rule.

so far i've simply taken logs and brought the power in front leaving me with $$ n\log \left(1+ \frac xn\right) = x $$

share|cite|improve this question
I think, I saw something like this before at the site. – Babak S. Dec 28 '12 at 11:08
@BabakSorouh. Yes definitley – Amr Dec 28 '12 at 11:08
@BabakSorouh do either of you remember the question title by any chance? – jill Dec 28 '12 at 11:10
Possible duplicate:… – Julian Kuelshammer Dec 28 '12 at 11:19
Sometimes this is the definition of $e^x$. What is your definition of $e^x$? – Henning Makholm Dec 28 '12 at 12:05

The ‘$=x$’ is getting ahead of yourself a bit. Let $$L=\lim_{n\to\infty}\left(1+\frac{x}n\right)^n\;,$$ and take the logarithm to get

$$\begin{align*} \ln L&=\ln\lim_{n\to\infty}\left(1+\frac{x}n\right)^n\\ &=\lim_{n\to\infty}\ln\left(1+\frac{x}n\right)^n\\ &=\lim_{n\to\infty}n\ln\left(1+\frac{x}n\right)\;, \end{align*}$$

where the interchange of the log and the limit is justified by the fact that the logarithm function is continuous.

This limit is now a so-called $\infty\cdot 0$ indeterminate form, and there is a standard approach to those: move one of the factors into the denominator. In this case we have

$$\ln L=\lim_{n\to\infty}\frac{\ln\left(1+\frac{x}n\right)}{1/n}\;,$$

a limit in which both numerator and denominator approach $0$ as $n\to\infty$. Now you can apply l’Hospital’s rule.

Don’t forget that at this point you’re actually finding $\ln L$, not $L$, so you’ll have to exponentiate to get $L$.

share|cite|improve this answer

Taking $\,n\,$ as a continuous variable:



Now just apply the exponential function at both ends of the above, remembering this function is a continuous one on the whole real line.

share|cite|improve this answer
@jill: Note that to ... this function is continuous one... above. It is very important for getting the desire result. – Babak S. Dec 28 '12 at 11:17

$$\lim_{n\to\infty} (1 + \frac xn)^n =\lim_{n\to\infty} e^{n\ln(1 + \frac xn)} $$ The limit $$\lim_{n\to\infty} n\ln(1 + \frac xn)=\lim_{n\to\infty} \frac{\ln(1 + \frac xn)}{\frac1n}=\lim_{n\to\infty} \frac{\frac{1}{1 + \frac xn}\frac{-x}{n^2}}{-\frac1{n^2}}=\lim_{n\to\infty} \frac{x}{1 + \frac xn}=x$$ By continuity of $e^x$, $$\lim_{n\to\infty} (1 + \frac xn)^n =\lim_{n\to\infty} e^{n\ln(1 + \frac xn)}=e^x $$

share|cite|improve this answer

If $\lim\limits_{x\to{+\infty}} f(x)^{g(x)}$ be as $1^{+\infty}$, which is an indeterminate form, then we have this fact that: $$\lim_{x\to{+\infty}} f(x)^{g(x)}=e^{\lim\limits_{x\to +\infty}\big(f(x)-1\big)g(x)}$$ Try to verify and then prove it. :)

share|cite|improve this answer

Write the limit in the following form $$\lim_{n\to\infty}\frac{\log(1+x/n)}{1/n}=\lim_{n\to\infty}\frac{f(n)}{g(n)}$$ where $f(n)=\log (1+x/n)$ and $g(n)=1/n$ and the limit is in $0/0$ form, so applying L'Hospitals rule we have the above limit is same as $$\lim_{n\to\infty}\frac{f^\prime(n)}{g^\prime(n)}=\lim_{n\to\infty}\frac{1/(1+x/n).(-x/n^2)}{-1/n^2}=\lim_{n\to\infty}\frac{x}{1+x/n}=x$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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