Proof that $\lim_{x\to\infty}\left(\frac{x}{x-1}\right)^x=e$. I was at work and I was curious if there was a limit for the formula
$$f(x)=\left(\frac{x}{x-1}\right)^2$$
And I became curious on Desmos to see if there would a different limit for
$$f(x)=\left(\frac{x}{x-1}\right)^x$$
It looked like it approached a constant value and, upon plugging it into a limit calculator, it returned
$$\lim_{x\to\infty}\left(\frac{x}{x-1}\right)^x=e$$
What is the proof of this? I managed to work it down to $x^x(x-1)^{-x}$, but my calculus isn't good enough for me to work it past there...

This is something you can remember when facing indetermination problems of the type $1^{\infty}$, it’s often used in many problems.
You have:
$$
\left(\frac{x}{x-1}\right)^x=e^{x \ln (\frac{x}{x-1})}
=e^{x \ln (1-\frac{1}{x-1})}$$
We know that when $X \rightarrow 0$, we have $\ln (1+X)$ equivalent to $X$. Therefore, when $x \rightarrow +\infty$, our expression is equivalent to $e^{\frac{x}{x-1}}$. You can now conclude that the limit of your expression is $e$, as you guessed it!
