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To prove

$$\lim_{x\rightarrow a} \frac{x^n-a^n}{x-a}=na^{n-1}$$

The proof is easy when we take $n$ as positive integer and $a$ any positive real number.

In my book it is given that the result is true even when $n$ is any rational number and $a$ any positive real number. But the proof is not given.

Please provide some hint to construct the proof for general case.

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$$\begin{array}{rcl} \displaystyle \lim_{x\rightarrow a} \frac{x^n-a^n}{x-a} &=& \displaystyle \lim_{x\rightarrow a} \frac{x^{p/q}-a^{p/q}}{x-a} \\ &=& \displaystyle \lim_{x\rightarrow a} \frac{\left(x^{p/q}-a^{p/q}\right)\left(x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}\right)}{\left(x-a\right) \left(x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}\right)} \\ &=& \displaystyle \lim_{x\rightarrow a} \frac{x^p - a^p}{\left(x-a\right) \left(x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}\right)} \\ &=& \displaystyle \lim_{x\rightarrow a} \frac{\left(x-a\right) \left(x^{p-1} + x^{p-2}a + \cdots + xa^{p-2} + a^{p-1}\right)}{\left(x-a\right) \left(x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}\right)} \\ &=& \displaystyle \lim_{x\rightarrow a} \frac{x^{p-1} + x^{p-2}a + \cdots + xa^{p-2} + a^{p-1}}{x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}} \\ &=& \displaystyle \frac{a^{p-1} + a^{p-2}a + \cdots + aa^{p-2} + a^{p-1}}{a^{(q-1)p/q} + a^{(q-2)p/q}a^{p/q} + \cdots + a^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}} \\ &=& \displaystyle \frac{pa^{p-1}}{qa^{(q-1)p/q}} \\ &=& \displaystyle \frac p q a^{(p-1)-(q-1)p/q} \\ &=& \displaystyle \frac p q a^{p-1-p+p/q} \\ &=& \displaystyle \frac p q a^{p/q-1} \\ &=& \displaystyle n a^{n-1} \\ \end{array}$$

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    $\begingroup$ Whether you realized or not the proof you gave assumes the continuity of function $x^{1/q}$ at $a$ and this is not as trivial as proving continuity for $x^{q} $. It requires essentially the theorem concerning continuity of inverse functions. $\endgroup$ – Paramanand Singh May 15 '17 at 12:18
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One possible way to show it.

Let $x=y+a$ to change the problem to $$\lim_{x\rightarrow a} \frac{x^n-a^n}{x-a}=\lim_{y\rightarrow o} \frac{(y+a)^n-a^n}{y}$$ Now, use the generalized binomial theorem or Taylor series around $y=0$ $$(y+a)^n=a^n+n y a^{n-1}+\frac{1}{2}n (n-1) y^2 a^{n-2}+O\left(y^3\right)$$ which makes $$A=\frac{(y+a)^n-a^n}{y}=\frac{a^n+n y a^{n-1}+\frac{1}{2}n (n-1) y^2 a^{n-2}+O\left(y^3\right) -a^n}{y}$$ that is to say $$A=n a^{n-1}+\frac{1}{2}n (n-1) y a^{n-2}+O\left(y^2\right)$$ which shows the limit and also how it is approached.

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  • $\begingroup$ Most proofs of generalized binomial theorem rely on the limit in question here. One can however prove the general binomial theorem without this limit also but with somewhat more effort. $\endgroup$ – Paramanand Singh May 15 '17 at 12:20
  • $\begingroup$ @ParamanandSingh. I hope that you will not be surprised if I say that I fully agree with you. This is why I started writing One possible way to show it. With this kind of problems, I always have the feeling that we are inside an infinite loop. Cheers and thahnks for your comment. $\endgroup$ – Claude Leibovici May 15 '17 at 17:19
  • $\begingroup$ Believe it or not, but I somehow missed your opening sentence. Pending +1 delivered! $\endgroup$ – Paramanand Singh May 16 '17 at 0:01
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If $n=p/q$, do the substitution $x=y^q$ (which is surely possible for $x>0$, with $y>0$; do the other cases separately) and $a=b^q$, so the limit becomes $$ \lim_{y\to b}\frac{y^p-b^p}{y^q-b^q}= \lim_{y\to b}\frac{y^p-b^p}{y-b}\frac{y-b}{y^q-b^q}= \frac{pa^{p-1}}{qa^{q-1}} $$ the last equality for the integer case.


I find such computations a waste of time. There's a much faster way to prove the result.

For the function $f(x)=x^{\alpha}$ (where $\alpha$ is an arbitrary real number) defined for $x>0$ we have $f(x)=\exp(\alpha\log x)$ and the chain rule says $$ f'(x)=\exp(\alpha\log x)\frac{\alpha}{x}=x^{\alpha}\frac{\alpha}{x}= \alpha x^{\alpha-1} $$ For rational $\alpha$ this might be defined also for $x\le0$, but it's easy to do the case with the help of symmetries.

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  • $\begingroup$ This particular theorem seems like one you'd be proving before defining derivatives, or at any rate before having the derivatives of exp and log available for use. $\endgroup$ – Gareth McCaughan May 15 '17 at 11:51
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    $\begingroup$ @GarethMcCaughan I've nothing against trying one's hand with $n=2/3$ or $n=3/2$. The way I showed reduces the problem to the integers and could be used for explaining substitutions in limits (with proper handling of the various situations), but risks to give a wrong idea about the nature of mathematics. See the most voted answer to get what I mean. $\endgroup$ – egreg May 15 '17 at 11:55
  • $\begingroup$ I couldn't agree more with your last comment. +1 for answer as well as comment. $\endgroup$ – Paramanand Singh May 15 '17 at 12:21
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As it is( 0/0) form apply L lopital rule
Differentiate you get

=a^n-1÷(a-1) as limit tends to a ( 0/0) form So differentiate with respect to a

=na^(n-1)

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    $\begingroup$ Welcome to StackExchange. Please use this tutorial to help you typeset Math nicely on this site, to make your answer more readable $\endgroup$ – lioness99a May 18 '17 at 13:02

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