How to show that $\lim_{x\to 0} (1+x)^{1/x}$ is same as $\lim_{x\to +\infty} (1+1/x)^{x}$?

Question

I am trying to calculate $\lim_{x\to 0}(1+x)^{1/x}$ by using that $\lim_{x\to+\infty}(1+1/x)^x=e$.

By substitution with $a=\frac1x$ I get that $\lim_{x\to 0}(1+x)^{1/x}= \lim_{a\to +-\infty}(1+\frac1a)^{a}$. So the Problem is that $x$ approaches $+\infty$ and $-\infty$ (not only $+\infty$).

Is there an easy way to show that $lim(1+a)^{1/a}$ is the same as $a$ approaches $-\infty$ and $+\infty$?

Answer 1

I would suggest to apply the following line of reasoning $$ \lim_{x\to-\infty}\left(1+\frac{1}{x}\right)^{1/x} = \lim_{x\to\infty}\left(1+\frac{1}{-x}\right)^{-1/x} = \frac1{\lim_{x\to\infty} \left(1+\frac{-1}{x}\right)^{1/x}} = \frac{1}{e^{-1}} = e. $$

I don't know if there is a better way of approaching this, but I hope this helps!

Answer 2

There is confusion in your question and the substitution is wrong.

In fact,

$$\lim_{x\to\infty}(1+x)^{1/x}=\lim_{x\to\infty}x^{1/x}=1$$

and

$$\lim_{x\to-\infty}(1+x)^{1/x}$$ is not defined as it involves real powers of negative.

There are no connections between

$$\lim_{x\to0}(1+x)^{1/x}=\lim_{x\to\infty}\left(1+\frac1x\right)^x=\lim_{x\to-\infty}\left(1+\frac1x\right)^x$$

and

$$\lim_{x\to\infty}(1+x)^{1/x}=\lim_{x\to0^+}\left(1+\frac1x\right)^x.$$