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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.

  • 0
    @GNUSupporter: And given that you were addressing a 7 year old edit, perhaps you could have taken the few seconds to see if your scolding was perhaps a bit past the statute of limitations. By the way: taking a seven-year-old post and **bumping** it to the front page with an edit that addresses nothing but style is likewise rather frowned upon. Its most recent prior edit was a year ago. Perhaps you might also consider that next time you decide to do edits without doing much follow-up or thinking about them.2018-03-19

5 Answers 5

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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.2011-04-30
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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.2011-04-30
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$\frac{x^{1/m}-1}{x^{1/n}-1} = \frac{e^{\log(x^{1/m})}-1}{e^{\log(x^{1/n})}-1} = \frac{e^{\frac1{m}\log(x)}-1}{e^{\frac1{n}\log(x)}-1} = \frac{e^{\frac1{m}\log(x)}-1}{\log(x)} \frac{\log(x)}{e^{\frac1{n}\log(x)}-1}$

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

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    @Nir: Yes. Thanks.2011-05-03