# Thread: Why does this limit equal e?

1. ## 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!

2. Because that's a definition of e.

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

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

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

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

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

use L'Hopital's rule ...

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

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

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

$\ln{y} = 1$

$y = e$

to check it yourself, graph the function $y = \left(1 + \frac{1}{x}\right)^x$ in your calculator and see what value the function approaches as x gets large.

4. 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 $L=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$, apply natural logs to both sides to get

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

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

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

Does this clarify things?

5. 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!

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### which limit is equal to e

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