prove that $\lim\limits_{x\to 1}\frac{x^{1/m}-1}{x^{1/n}-1}=\frac{n}{m}$

Question

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.

Answer 1

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.

Answer 2

Are you aware of L'Hôpital's rule? It is useful in evaluating the limits of fractions such as yours.

Answer 3

One thing about limits is that, if they exist, the "speed" at which you approach them doesn't matter. That is to say, $\lim_{x\rightarrow 1}\frac{x^{1/m}-1}{x^{1/n}-1} = \lim_{x^{1/n}\rightarrow 1}\frac{x^{1/m}-1}{x^{1/n}-1} = \lim_{y\rightarrow 1}\frac{y^{n/m}-1}{y-1}$. If you then apply L'Hopital's rule, you should get your answer.

Answer 4

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 ${\displaystyle {{1 \over m} \over {1 \over n}} = {n \over m}}$.

Answer 5

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

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