# Why does this limit equal e?

• Nov 29th 2009, 03:52 PM
ilovecalculus
Why does this limit equal e?
In several examples in my calculus class we were going over sums, using limit tests, and he simplified it down to something like:

(1+1/n)^n as n->00 (infinity)

and he said that equals e (2.71...) and nobody else seemed to question it but I really don't understand how that makes sense. Could someone explain this to me? Thank you!
• Nov 29th 2009, 04:19 PM
MathTooHard
Because that's a definition of e.
• Nov 29th 2009, 04:28 PM
skeeter
Quote:

Originally Posted by ilovecalculus
In several examples in my calculus class we were going over sums, using limit tests, and he simplified it down to something like:

(1+1/n)^n as n->00 (infinity)

and he said that equals e (2.71...) and nobody else seemed to question it but I really don't understand how that makes sense. Could someone explain this to me? Thank you!

there may be a reason he didn't show you why. hope you can follow this ...

let $\displaystyle y = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$

also note that $\displaystyle 1 + \frac{1}{n} = \frac{n+1}{n}$ ...

$\displaystyle \ln{y} = \lim_{n \to \infty} \ln\left(\frac{n+1}{n}\right)^n$

$\displaystyle \ln{y} = \lim_{n \to \infty} n \ln\left(\frac{n+1}{n}\right)$

$\displaystyle \ln{y} = \lim_{n \to \infty} \frac{\ln(n+1) - \ln{n}}{\frac{1}{n}}$

use L'Hopital's rule ...

$\displaystyle \ln{y} = \lim_{n \to \infty} \frac{\frac{1}{n+1} - \frac{1}{n}}{-\frac{1}{n^2}}$

$\displaystyle \ln{y} = \lim_{n \to \infty} \frac{-\frac{1}{n(n+1)}}{-\frac{1}{n^2}}$

$\displaystyle \ln{y} = \lim_{n \to \infty} \frac{n}{n+1}$

$\displaystyle \ln{y} = 1$

$\displaystyle y = e$

to check it yourself, graph the function $\displaystyle y = \left(1 + \frac{1}{x}\right)^x$ in your calculator and see what value the function approaches as x gets large.
• Nov 29th 2009, 04:31 PM
Chris L T521
Quote:

Originally Posted by ilovecalculus
In several examples in my calculus class we were going over sums, using limit tests, and he simplified it down to something like:

(1+1/n)^n as n->00 (infinity)

and he said that equals e (2.71...) and nobody else seemed to question it but I really don't understand how that makes sense. Could someone explain this to me? Thank you!

It can be proved using L'Hopital's rule. If you set $\displaystyle L=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$, apply natural logs to both sides to get

$\displaystyle \ln L=\lim_{n\to\infty}\frac{\ln\!\left(1+\frac{1}{n}\ right)}{\frac{1}{n}}$.

It turns out that $\displaystyle \ln L=1\implies L=e^1=e$.

So we can say that $\displaystyle \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e$.

Does this clarify things?
• Nov 29th 2009, 04:37 PM
ilovecalculus
Thanks to everyone! It makes a lot more sense now, and I guess it's just something I'll have to remember for future reference!