Prove that lim...=e

**Runty** · February 3rd 2010, 03:00 PM

Prove that

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

Hints outlined for this question are as follows.

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

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

**Plato** · February 3rd 2010, 03:19 PM

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

**Runty** · February 3rd 2010, 03:31 PM

Originally Posted by Plato

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

Didn't quite get what you were getting at at first, but then I realized it. Thanks.

**Drexel28** · February 3rd 2010, 05:53 PM

Originally Posted by Runty

Prove that

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

Hints outlined for this question are as follows.

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

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

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

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