Why is $1^{\infty}$ undefined when the limit of it is $\operatorname e$? What I mean to say is, we know that:
$$\lim_{x\to\infty} \biggl(1+\frac{1}{x}\biggr)^x = e$$
But,
$$\lim_{x\to\infty} \biggl(\frac{1}{x}\biggr) = 0$$
Thus,
$$\lim_{x\to\infty} (1+0)^x = e$$
Finally, $1^{\infty} = e$.
Why is this incorrect? Can anyone explain where the math is wrong?
 A: Fundamentally, the problem is that you're treating $\infty$ as though it's a real number when it's really not. 
A: $1^\infty$ is indeterminate form. So $1^\infty=e$ may be true, but it may also happen that $1^\infty\ne e$
For example take $\alpha>0$ a real number and write:
$$\lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right)^{\alpha x}=e^\alpha$$
So you might as well conclude that $1^\infty=e^\alpha$
A: When we state that $1^\infty$ is an indeterminate form, we mean that if $f$, $g$ are two real valued functions such that $x_0$ is in the intersection of their domains and
$$\lim_{x\to x_0}f(x)=1 \qquad \lim_{x\to x_0}g(x)=+\infty \tag{1}$$
then the limit
$$\lim_{x\to x_0}f(x)^{g(x)}$$
is not uniquely defined by the condition $(1)$.

In the following example, we show that for every $a>0$ there exist functions  $f_a, g_a$ and $x_0\in \mathbb {\tilde R}$ satisfying $(1)$ such that
$$\lim_{x\to x_0}f_a(x)^{g_a(x)}=a$$

Take $a>0$ and define
  $$f_a(x)=\left (1+\frac {\log(a)}{x}\right )\qquad g(x)=x$$
  As you can see $f_a$ and $g$ satisfy $(1)$ in $x_0=+\infty$ for every $a$, because
  $$\lim_{x\to x_0}f_a(x)^{g(x)}=\lim_{x\to \infty}\left(1+\frac {\log(a)}{x} \right)^x"="1^\infty$$
  Although, taking the limit,
  $$\lim_{x\to x_0}f_a(x)^{g(x)}=\lim_{x\to \infty}\left(1+\frac {\log(a)}{x} \right)^x=e^{\log(a)}=a$$
  so the limit of an indeterminate form $1^\infty$ can be any positive real number.

A: There is a flow in your logic, you separated the limit then you did it part by part.
This is correct on it's own:
$\lim_{x\to\infty} (1+\frac{1}{x})^x = e$
This is correct on it's own as well $\lim_{x\to\infty} (\frac{1}{x}) = 0$
This part is nonesensical jumoing to conclusion:
Thus,
$$\lim_{x\to\infty} (1+0)^x = e$$
because the correct version is this $\lim_{x\to\infty} (1+0)^x = 1$
Finally, $1^{\infty} = e$. Incorrect conclusion
Why is this incorrect? Can anyone explain where the math is wrong?
Math is not wrong, the logic is wrong, limit of a function is not same as function of the limit $(1+\frac{1}{x})^x$ is a function but you are applying the limit to some parts first then to the other parts, limit must be applied the whole, not part by part.
A: It's not necessarily incorrect.  Notice that if we take the log of $1^\infty$, we have
$$\ln1^\infty=\infty\ln1=\infty\times0$$
And we all know very well that this indeterminate form can equal anything.
However, you reasoning is still wrong, since one may not move limits around like that.
A: It's a common mistake to do a limit inside a limit like that. Yes, $\lim_{x \to \infty}\frac{1}{x} = 0$, but that tells you nothing about $\lim_{x \to \infty}(1 + \frac{1}{x})^x$ - for example, $\lim_{x\to\infty}x\cdot\frac{1}{x} = \lim_{x\to\infty}\frac{x}{x} = \lim_{x\to\infty}1 = 1$, even though it looks like you can say $\lim_{x\to\infty}x\cdot\frac{1}{x} = \lim_{x\to\infty}x\cdot 0 = \lim_{x\to\infty}0 = 0$. The key thing is that saying something like "$\lim_{x \to \infty}\frac{1}{x} = 0$" isn't saying "$\frac{1}{x}$ eventually becomes zero", it's saying "$\frac{1}{x}$ eventually gets as close as you like to zero". So, for large enough $x$, $1 + \frac{1}{x}$ is very close to $1$. But when $x$ is enormous, taking a number very close to $1$ and raising it to the $x$th power takes that "closeness" and pulls it wide open - $1.0001^{100000000}$ is so large Google refuses to calculate it. It so happens that the "smallness" of $\frac{1}{x}$ and the "bigness" of raising something to the $x$th power balance out in just the right way so that $\lim_{x \to \infty}(1 + \frac{1}{x})^x = e$. But change anything and you upset the balance - $\lim_{x \to \infty}(1 + \frac{1}{2x})^x = \sqrt{e}$, while $\lim_{x \to \infty}(1 + \frac{1}{x})^{2x} = e^2$.
The point is, you can never just replace something inside a limit with its limit, unless you're deleting the limit. That's an extremely dangerous operation which will only yield the right answer if you got obscenely lucky - because when you do that, you erase all of the information about how fast the expression approaches its limit, so you sacrifice the opportunity to balance things out.
