# prove that $\displaystyle\lim_{x\rightarrow 1}\frac{x^{1/m}-1}{x^{1/n}-1}=\frac{n}{m}$

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.

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I suggest playing around with various inequalities and using the squeeze theorem. For example, the inequality $(1+\epsilon)^a\leq 1+a\epsilon$ when $a,\epsilon >0$, helps quite a bit. – Grumpy Parsnip Apr 30 '11 at 21:19
Like Jim said, rather than the accepted answer, a more analytical way of thinking this would be using Taylor expansion: writing $x^{1/m} = (1 + \Delta x)^{1/m} = 1 + \frac{1}{m} \Delta x + o(\Delta x^2)$, the higher order terms will disappear when letting $\Delta x \rightarrow 0$ in that quotient you have. – Shuhao Cao May 3 '11 at 17:00

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.

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

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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.

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If you're going to apply L'Hopital, why not do it immediately? But instead of invoking L'Hopital in your last step, one can observe that your transformed limit is a derivative at 1, see Bill Dubuque's answer. – wildildildlife Apr 30 '11 at 21:08

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

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Yeah, I can't use it in this stage. I need to teach this someone who can't use LHopital right now. – user6163 Apr 30 '11 at 20:51

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

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

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isn't this a really indirect way of saying "use $lim_{\epsilon\rightarrow 0} (1+\epsilon)^{1/m} = 1+(\epsilon/m)+O(\epsilon^2)$ – samanwita May 1 '11 at 3:41
@samanwita: I have used the fact $\lim_{y \rightarrow 0} \frac{e^y-1}{y} = 1$ which is probably one of the first limits people are taught. – user17762 May 1 '11 at 4:16
Did you mean that $x$ tends to 1 in the second row? – user6163 May 3 '11 at 9:45
@Nir: Yes. Thanks. – user17762 May 3 '11 at 16:36