# Prove that lim...=e

• Feb 3rd 2010, 03:00 PM
Runty
Prove that lim...=e
Prove that

$\displaystyle \lim_{h \to 0}(1+h)^{1/h}=e$

Hints outlined for this question are as follows.

Since$\displaystyle \frac{d}{dx}\ln1=1$,
As $\displaystyle h \to 0$, $\displaystyle \frac{\ln(1+h)-\ln1}{h}=\frac{\ln(1+h)}{h} \to 1$
A second hint is that if $\displaystyle g$ is continuous at $\displaystyle c$ and $\displaystyle f$ is continuous at $\displaystyle g(c)$, then $\displaystyle f\circ g$ is continuous at $\displaystyle c$.

You are not allowed to use L'Hospital's Rule to solve this question. This is one of the requirements, I'm afraid.
• Feb 3rd 2010, 03:19 PM
Plato
Do you understand that $\displaystyle \lim _{n \to \infty } \left( {1 + \frac{1}{n}} \right)^n = e$?
If so let $\displaystyle h=\frac{1}{n}$.
• Feb 3rd 2010, 03:31 PM
Runty
Quote:

Originally Posted by Plato
Do you understand that $\displaystyle \lim _{n \to \infty } \left( {1 + \frac{1}{n}} \right)^n = e$?
If so let $\displaystyle h=\frac{1}{n}$.

Didn't quite get what you were getting at at first, but then I realized it. Thanks.
• Feb 3rd 2010, 05:53 PM
Drexel28
Quote:

Originally Posted by Runty
Prove that

$\displaystyle \lim_{h \to 0}(1+h)^{1/h}=e$

Hints outlined for this question are as follows.

Since$\displaystyle \frac{d}{dx}\ln1=1$,
As $\displaystyle h \to 0$, $\displaystyle \frac{\ln(1+h)-\ln1}{h}=\frac{\ln(1+h)}{h} \to 1$
A second hint is that if $\displaystyle g$ is continuous at $\displaystyle c$ and $\displaystyle f$ is continuous at $\displaystyle g(c)$, then $\displaystyle f\circ g$ is continuous at $\displaystyle c$.

You are not allowed to use L'Hospital's Rule to solve this question. This is one of the requirements, I'm afraid.

Let $\displaystyle 1=\lim_{x\to0}\frac{\ln(1+x)}{x}=\lim_{x\to0}\ln\l eft(\left(1+x\right)^{\frac{1}{x}}\right)=\ln\left (\lim_{x\to0}\left(1+x\right)^{\frac{1}{x}}\right)$. The last part follows from the hint stuff. The conclusion follows by exponentiation.