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

Solution 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.

Solution 2:

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

Solution 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.