TITLE: prove that $\lim\limits_{x\to 1}\frac{x^{1/m}-1}{x^{1/n}-1}=\frac{n}{m}$
QUESTION [14 upvotes]: I'm trying to find a way to prove this:
EDIT: without using LHopital theorem.
$$\lim_{x\rightarrow 1}\frac{x^{1/m}-1}{x^{1/n}-1}=\frac{n}{m}.$$
Honestly, I didn't come with any good idea.
We know that $\lim_{x\rightarrow 1}x^{1/m}$ is $1$.
I'd love your help with this.
Thank you.

REPLY [6 votes]: You can rewrite the limit as
$$\lim_{x \rightarrow 1} {{x^{1\over m} - 1 \over x - 1} \over {x^{1 \over n} - 1 \over x- 1}}$$
By the quotient rule for limits this is exactly
$${\lim_{x \rightarrow 1} {x^{1 \over m} - 1 \over x - 1} \over \lim_{x \rightarrow 1} {x^{1 \over n} - 1 \over x - 1}}$$
But notice that for any $\alpha$, ${\displaystyle \lim_{x \rightarrow 1} {x^{\alpha} - 1 \over x - 1}}$ is just the limit of difference quotients giving the definition of the derivative of the function $x^{\alpha}$ when evaluated at $x = 1$. So the limit is $\alpha$. So the limit in this question will be ${\displaystyle {{1 \over m} \over {1 \over n}} = {n \over m}}$.