prove that $\lim\limits_{x\to 1}\frac{x^{1/m}-1}{x^{1/n}-1}=\frac{n}{m}$ 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.
HINT $\ $ If you change variables $\rm\ z = x^{1/n} $ then the limit reduces to a very simple first derivative calculation. See also some of my prior posts for  further examples of limits that may be calculated simply as first derivatives.
$$\frac{x^{1/m}-1}{x^{1/n}-1} = \frac{e^{\log(x^{1/m})}-1}{e^{\log(x^{1/n})}-1} = \frac{e^{\frac1{m}\log(x)}-1}{e^{\frac1{n}\log(x)}-1} = \frac{e^{\frac1{m}\log(x)}-1}{\log(x)} \frac{\log(x)}{e^{\frac1{n}\log(x)}-1}$$

$$\lim_{x \rightarrow 1} \frac{x^{1/m}-1}{x^{1/n}-1} = \lim_{\log(x) \rightarrow 0} \frac{e^{\frac1{m}\log(x)}-1}{\log(x)} \frac{\log(x)}{e^{\frac1{n}\log(x)}-1} = \lim_{y \rightarrow 0} \frac{e^{\frac{y}{m}}-1}{y} \frac{y}{e^{\frac{y}{n}}-1} = \frac1{m} \frac1{\frac1{n}} = \frac{n}{m}$$