# L'Hospital's Rule Question.

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 $$n\log \left(1+ \frac xn\right) = x$$

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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: math.stackexchange.com/questions/39170/… – 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 $L$.

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

$$\lim_{n\to\infty}\log\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}n\log\left(1+\frac{x}{n}\right)=\lim_{n\to\infty}\frac{\log\left(1+\frac{x}{n}\right)}{\frac{1}{n}}\stackrel{\text{L'Hospital}}=$$

$$=\lim_{n\to\infty}-\frac{x}{n^2}\frac{\frac{1}{1+\frac{x}{n}}}{-\frac{1}{n^2}}=x\lim_{n\to\infty}\frac{1}{1+\frac{x}{n}}=x$$

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

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@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 $e^x$, $$\lim_{n\to\infty} (1 + \frac xn)^n =\lim_{n\to\infty} e^{n\ln(1 + \frac xn)}=e^x$$

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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. :)

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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 $$\lim_{n\to\infty}\frac{f^\prime(n)}{g^\prime(n)}=\lim_{n\to\infty}\frac{1/(1+x/n).(-x/n^2)}{-1/n^2}=\lim_{n\to\infty}\frac{x}{1+x/n}=x$$

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