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...
 A: If you know that $$\lim_{x \to \infty} \left( 1 + \frac tx \right)^x = e^t$$ for any real $t$ you can use the particular choice $t = -1$ to find that $$\lim_{x \to \infty} \left(\frac{x-1}{x} \right)^x = \lim_{x \to \infty} \left( 1 - \frac 1x \right)^x = e^{-1} = \frac 1e.$$
Now take reciprocals and use an appropriate limit law.
A: 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!
A: Let $\log$ denote the natural logarithm.
For any $x>1$ we have, $$\dfrac{x}{x-1}>0$$ therefore,
$$\forall x>1,\left(\dfrac{x}{x-1} \right)^x= \mathrm{e}^{x\log\left(\frac{x}{x-1}\right)}=\mathrm{e}^{-x \log\left(\frac{x-1}{x}\right)}=e^{-x\log \left(1-\frac{1}{x}\right)}$$
Moreover,
$$\log\left(1-\frac{1}{x}\right) \underset{+\infty}{\sim}-\frac{1}{x}$$
Therefore,
$$-x \log\left(1-\frac{1}{x}\right) \underset{x\rightarrow +\infty}{\longrightarrow} 1$$
Hence, using the fact $\exp$ is continuous,
$$\boxed{\lim_{x \rightarrow+\infty} \left(\dfrac{x}{x-1} \right)^x = e^1}$$
