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

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)

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

• Thank you, is there a proof of why it's legitimate to make the substitution $t = x-1$ and then say that $\lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)^{x-1} = \lim\limits_{t \to \infty}\left(\dfrac{t+1}{t}\right)^{t}$? I almost wrote it the same way in my OP and then realized I wouldn't be sure how to prove the legitimacy of that substitution. – jeremy radcliff Nov 30 '15 at 23:41
• @jeremyradcliff Consider the definition of $\lim\limits_{x \to \infty}f(x) = L \neq \pm \infty$: for all $\epsilon > 0$ there is a $M > 0$ such that for all $x > M$, $|f(x) - L| < \epsilon$. Now we know $\lim\limits_{t \to \infty}\left(\dfrac{t+1}{t}\right)^{t} = e$. Let $f(t) = \left(\dfrac{t+1}{t}\right)^{t}$. So for all $\epsilon > 0$, there is a $M > 0$ such that for all $t > M$, $|f(t) - e| < \epsilon$. I claim that $\lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)^{x-1} = e$. Let $\left(\dfrac{x}{x-1}\right)^{x-1} = g(x)$. (continued)... – Clarinetist Dec 1 '15 at 0:06
• @jeremyradcliff I need to show that for all $\epsilon > 0$ there is a $N> 0$ ($N$ not necessarily equal to $M$) such that for all $x > N$, $|g(x) - e| < \epsilon$. So I need to find $N$. Observe that $g(x) = f(x-1)$. We know that for all $t > M$, $|f(t) - e| < \epsilon$. Take $N = M-1$. (Do you see why?) Then for all $x > M - 1$, $|g(x) - e| < \epsilon$. – Clarinetist Dec 1 '15 at 0:13
• Thank you for taking the time to explain, this is very helpful. – jeremy radcliff Dec 1 '15 at 1:09
• @jeremyradcliff: The substitution $t = x - 1$ is guaranteed by the general rule of substitution of limits. See a related answer math.stackexchange.com/a/1073047/72031 which explains this rule. Also it is easy to formulate this rule for the case when the limit variable tends to $\infty$ instead of a finite number. – Paramanand Singh Dec 1 '15 at 12:57

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

$$\left(\frac{x}{x-1}\right)^x=\left(\frac{x-1+1}{x-1}\right)^x=\left(1+\frac{1}{x-1}\right)^x=\left[\left(1+\frac{1}{x-1}\right)^{x-1}\right]^{\frac{x}{x-1}}$$ 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 $e$.

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 $t\mapsto\log(1+at)$.