# Limit is equivalent

1. Dec 22, 2004

### quasar987

I don't understand how to show that

$$\lim_{n \rightarrow \infty} \left(1-\frac{a}{n} \left)^{n} = e^{-a} \ \ \forall a \in \mathbb{R}$$

For exemple, if I say "Let x be the real number such that $n=-ax \Leftrightarrow x=-n/a$. Then the limit is equivalent to

$$\lim_{-ax \rightarrow \infty} \left(1+\frac{1}{x} \right)^{-ax} = \left(\lim_{-ax \rightarrow \infty} \left(1+\frac{1}{x} \right)^{x} \right)^{-a}$$

"but $-ax \rightarrow \infty$ is not equivalent to $x \rightarrow \infty$, so I can't conclude that the limit is $e^{-a}$.

What am I missing here ?

2. Dec 22, 2004

### StatusX

a is constant. true, x will either be going to negative or positive infinity depending on the sign of a, but the definition of e works for either:

let: $$u = -x$$

$$e = \lim_{x \rightarrow \infty} (1+\frac{1}{x} )^{x}$$

$$= \lim_{-u \rightarrow \infty} (1+\frac{1}{-u} )^{-u}$$

$$= \lim_{u \rightarrow -\infty} (\frac{1}{(1-\frac{1}{u})})^{u}$$

$$= \lim_{u \rightarrow -\infty} (\frac{1+\frac{1}{u}}{(1-\frac{1}{u^2})})^{u}$$

and the 1/u2 term becomes negligible, giving the result:

$$e = \lim_{u \rightarrow -\infty} (1+\frac{1}{u} )^{u}$$

edit: that may not be rigorous enough. you can show the bottom of the fraction above goes to 1 by taking the ln and using l'hopitals. in fact, you might want to just do that from the start.

Last edited: Dec 22, 2004
3. Dec 22, 2004

### Galileo

$$\lim_{n \to -\infty}\left(1+\frac{1}{n}\right)^n=e$$ is true too.

4. Dec 22, 2004

### Popey

Shouldn't -ax be an integer anyway?

Ok, the limit of (1+1/n)^n is e when n ->oo and n is an integer

but what happens when we get the value
(1+1/x)^x
where x is a very large real but not an integer ??

Shouln't we prove this cases ?

5. Dec 22, 2004

### matt grime

The identity e = lim(1+1/x)^x is true whether x is an integer or not (as you implicitly use yourself).

6. Dec 23, 2004

### dextercioby

Elegance is a quality of mathematics:
$$\lim_{n\rightarrow +\infty}(1-\frac{a}{n})^{n}=[\lim_{n\rightarrow +\infty}(1+\frac{1}{\frac{n}{-a}})^{\frac{n}{-a}}]^{-a}=[\lim_{\frac{n}{-a}\rightarrow\pm\infty}(1+\frac{1}{\frac{n}{-a}})^{\frac{n}{-a}}]^{-a}=e^{-a}$$
$$\lim_{n\rightarrow\pm\infty}(1+\frac{1}{n})^{n}=e$$

Daniel.

PS.The sign of "a" is irrelevant.It's important for it not to be "0".