# Is $\lim_{x\to\infty}(1+\frac{1}{x})^{z}$ always $e$? Why?

I see alot of expressions like $$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{z}=e$$ Where $z$ is an expression in $x$. Can $z$ be a constant tho?

But why is that? Also what if I have $$\lim_{x\to\infty}\left(1+\frac{1}{x+3}\right)^{z}$$

-
If $z$ is a constant, then your limit is $1$... – J. M. Apr 20 '12 at 6:59
@J.M. ... its very stupid of me not to see that ... maths is really not my thing ... anyways, what about the 2nd case? I guess its still $e$ since, the constant doesn't matter since $x\rightarrow\infty$? – Jiew Meng Apr 20 '12 at 7:03

No, if $z=2x$, the limit is $e^2$. If $z=x^2$, the limit doesn't exist (or, if you prefer, is infinite). If $z=\sqrt{x}$ the limit is $1$. And so on!
For your second question, similar remarks apply, but if $z=x$, the limit will be $e$. You can show this by observing that $$\left(1+\frac{1}{x+3}\right)^x=\left(1+\frac{1}{x+3}\right)^{x+3}\left(1+\frac{1}{x+3}\right)^{-3}.$$
It matters very heavily what $z$ actually is. For example, if it's the case that $z = 2x$, so that we have $(1 + \frac{1}{x})^{2x}$, this limit is $e^2$. This shouldn't be too surprising, as every term is exactly the square of the 'typical' limit.
If $z$ is a constant, then the limit is $1$. The fact that in the 'typical' limit we are raising everything to a really big power is very important. It's a bit nonintuitive at first, maybe, that taking a power of something that, for large $n$, is essentially $1$ can not be $1$, right? (I remember thinking that).