Why does $\lim_{x \to \infty} \big(1 + \frac{1}{x}\big)^x = \lim_{x \to 0} \big(1 + x\big)^{\frac{1}{x}}$?

Question

Is there a way to make sense of that relationship? Could you derive one from the other algebraically?

It looks like the first limit approaches $e$ from lower values (for positive $x$ values) whereas the second limit approaches $e$ from greater values, but i'm not sure what to make of that.

(I included the calculus tag because I've only seen the second limit in the context of trying to find the derivative of $\log(x)$)

Answer 1

You could start with the following substitution:$$t=\frac1x$$Putting this in the first equation gives us $$(1+t)^{1/t}$$

We also have to be careful with the limit, noting$$\lim_{x\to\infty}\frac1x=\lim_{t\to0^+}t$$Using this, we deduce that after substitution, our limit should come out to say $$\lim_{t\to0^+}(1+t)^{1/t}$$

Answer 2

The complex case
$$ (1+x)^{1/x} = \exp\left(\frac{\log(1+x)}{x}\right) $$ by definition. For complex $x$ with $|x|<1$, the principal value of log has expansion $$ \log(1+x) = x - \frac{x^2}{2} + \cdots $$ a power series with radius of convergence $1$. Then $$ \frac{\log(1+x)}{x} = 1 - \frac{x}{2} + \cdots $$ also a power series with radius of convergence $1$. The complex limit $$ \lim_{x \to 0} \frac{\log(1+x)}{x} = 1 $$ exists, and thus the complex limit $$ \lim_{x \to 0}\;(1+x)^{1/x} = \exp(1) $$ also exists.

The substitution in the other answers shows that the complex limit $$ \lim_{z \to \infty}\;\left(1+\frac{1}{z}\right)^z = \exp(1) $$ also.