Extend ${\bigl(1+\frac1x\bigr)}^{{x}}$ to $\overline{\mathbb R}$

Question

We can extend these functions to $\overline{\mathbb R}$ by taking limits says here.

\begin{align} \mathrm e^{-\infty} &= 0 \\ \mathrm e^{+\infty} &= \infty \\ \ln{\left|0\right|} &= -\infty \\ \ln{\left|\pm\infty\right|} = +\infty \end{align}

Then I am going to extend ${\bigl(1+\frac1x\bigr)}^{{x}}$ to $\overline{{\mathbb R}}$, if according to the Arithmetic operations defined here, I should get $\frac{{1}}{\infty}={0}$, then I think I should extend the function to ${\left({1}+\frac{{1}}{\infty}\right)}^{\infty}={\left({1}+0\right)}^{\infty}={1‌​}$, but according to the limit , I think I should get ${\left({1}+\frac{{1}}{\infty}\right)}^{\infty}={\mathrm e}$ ,so what's wrong here ? Which cases the difference ? Which is right ?

Answer 1

Like KittyL said in the comments, it's wrong to say that it equals $1^\infty$. It depends on what $x$ you look at first. If you take a look at the inner x first you indeed get that $\frac{1}{\infty}=0$ so that $1+\frac{1}{\infty}=1$ and $1^\infty=1$. But if you take a look at the $x$ in the exponent first, you see that $1+\frac{1}{x} > 1, \forall x>0$ so $(1+\frac{1}{x})^\infty = \infty$ no matter what $x$ you fill in.

Both solutions are clearly wrong. You have to look at both $x$'s at the same time. This will give you $e$ as the result.