Prove that $\lim_{x \to \infty}\big(\frac{x}{x-1}\big)^x$ is also $e$.

Question

Trying to make sense out of the idea that $100\%$ continuous decay is $\frac{1}{e}$, I thought about this:

You can express $1+\frac{1}{x}$ as $\frac{x+1}{x}$, such that $\big(1+\frac{1}{x}\big)^x = \big(\frac{x+1}{x}\big)^x$

And you can express $1-\frac{1}{x}$ as $\frac{x-1}{x}$, such that $\big(1-\frac{1}{x}\big)^x = \big(\frac{x-1}{x}\big)^x =\frac{1}{\big(\frac{x}{x-1}\big)^x}$

Now $\big(\frac{x}{x-1}\big)^x$ and $\big(\frac{x+1}{x}\big)^x$ look very similar, and I can imagine (and see on Mathematica) that $\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x$ also approaches $e$.

However I'm clueless about limit proofs, how do you prove that?

EDIT: Just had a last second insight. Can you say that $$\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x = \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot \lim_{x \to \infty} \big(\frac{x}{x-1}\big)$$ $$= \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot 1$$ $$=\lim_{x \to \infty} \big(\frac{x+1}{x}\big)^x = e ?$$

(not sure about that last limit transition...I mean I'm pretty sure it's true, just not sure how to write it; I would appreciate feedback on that, thank you)

Answer 1

For your second insight...

$$\lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)^{x}$$ You can split the limit into product terms if and only if the product terms both exist.

i.e., before splitting into a product, observe that $$\lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)^{x-1} = \lim\limits_{t \to \infty}\left(\dfrac{t+1}{t}\right)^{t}$$ using the substitution $t = x - 1$ (and as $x \to \infty$, intuitively, $t \to \infty$ as well). Obviously, this is equal to $e$.

Also observe that $\lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right) = 1$.

Because the product limits exist, $$\lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)^{x} = \left[ \lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)^{x-1}\right]\left[ \lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)\right] = e \cdot 1 = e\text{.}$$

Answer 2

$$ \lim_{n\to \infty} \left(\frac{n}{n-1}\right)^n=\lim_{n\to \infty} \left(1+\frac{1}{n-1}\right)^{n-1} \cdot \left(1+\frac{1}{n-1}\right)= e. $$

Answer 3

$$\left(\frac{x}{x-1}\right)^x=\left(\frac{x-1+1}{x-1}\right)^x=\left(1+\frac{1}{x-1}\right)^x=\left[\left(1+\frac{1}{x-1}\right)^{x-1}\right]^{\frac{x}{x-1}}$$ When $x\to \infty$, the exponent $\frac{x}{x-1}=\frac{1}{1-1/x}$ tends to $1$ and $\left(1+\frac{1}{x-1}\right)^{x-1}$ it is known tends to $e$.

Answer 4

If you know that $$ \lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{x}=e^a $$ then $$ \lim_{x\to\infty}\left(1-\frac{1}{x}\right)^{x}=e^{-1} $$ and so $$ \lim_{x\to\infty}\left(\frac{x}{1-x}\right)^x= \lim_{x\to\infty}\frac{1}{\left(\dfrac{x-1}{x}\right)^x}= \lim_{x\to\infty}\frac{1}{\left(1-\dfrac{1}{x}\right)^x}= (e^{-1})^{-1}=e $$ The proof of the first limit is easy: $$ \lim_{x\to\infty}\log\left(1+\frac{a}{x}\right)^{x}= \lim_{x\to\infty}x\log\left(1+\frac{a}{x}\right)= \lim_{t\to0^+}\frac{\log(1+at)}{t}=a $$ because it's the derivative at $0$ of the function $t\mapsto\log(1+at)$.