3

Please note that this question was asked by one of my students who doesn't know differentiation yet nor Lhopital nor mean value theorems. We teach limits before all these topics like differentiation , MVT , Lhopital , etc

$$\lim_{ x \to a} \frac{x^n-a^n}{x-a}=n\cdot a^{n-1}$$

I can prove this result for $n \in \mathbb Z$

And for $n \in \mathbb Q $ , that is when $n =\frac{p}{q}$ , I can prove the result using the result for $n \in\mathbb Z$.

But my question is this :

Since $\mathbb Z \subset \mathbb Q$ , why can't we prove this result only for $n \in \mathbb Q$ ?

Is there a method to prove $$\lim_{ x \to a} \frac{x^\frac{p}{q}-a^\frac{p}{q}}{x-a}=\frac{p}{q}\cdot a^{\frac{p}{q}-1}$$ without the result for $n \in \mathbb Z $ ?

Community
  • 1
Angelo Mark
  • 5,954
  • Lhopital easily seems to justify your claim. –  Sep 27 '16 at 17:00
  • 1
    @Avi Without using Lhopital ! – Angelo Mark Sep 27 '16 at 17:01
  • 1
    How can we prove any thing about anything to do with $x^{\frac pq}$ without appealing to the definition osf $x^{\frac pq} = \sqrt[q] x^p; p,q \in \mathbb Z$? – fleablood Sep 27 '16 at 17:01
  • 1
    General binomial theorem may be useful. – gcy-rolle Sep 27 '16 at 17:15
  • Or use mean value theorem for $f(x)=x^{\frac{p}{q}}$ – rtybase Sep 27 '16 at 17:24
  • 1
    "General binomial theorem may be useful" what is the general binomial theorem on rational exponents? – fleablood Sep 27 '16 at 17:27
  • Just take n= p/q. The result will follow from the general binomial theorem as n is non negative. ( just took gcy-rolle's hint half step further :)) –  Sep 27 '16 at 17:39
  • 1
    You might want to refer to https://en.m.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalised_binomial_theorem –  Sep 27 '16 at 17:41
  • @gcy-rolle, you are right about general binomial theorem, it leads to simple calculation of the limit, but it really is a Taylor expansion, and I doubt power series are studied before differentiation. Though, I guess OP could easily demonstrate calculation if the student would be willing to make "a leap of faith" that everything behaves just the same as in the case where $n$ is positive integer. – Ennar Sep 27 '16 at 20:03

7 Answers7

6

I'll prove it for $a=1$, it generalises fairly nicely from there:

$$\frac{x^{p/q}-1}{x-1}=\frac{(x^{1/q})^p-1}{(x^{1/q})^q-1}=\frac{\frac{(x^{1/q})^p-1}{x^{1/q}-1}}{\frac{(x^{1/q})^q-1}{x^{1/q}-1}}=\frac{1+x^{1/q}+x^{2/q}+\cdots+x^{p-1/q}}{1+x^{1/q}+x^{2/q}+\cdots+x^{q-1/q}}\to\frac{p}{q} \text{ as }x\to1$$

Note that the third equals sign comes from $\frac{y^n-1}{y-1}=y^{n-1}+y^{n-2}+\cdots+y+1$.

πr8
  • 10,800
  • "But my question is this : Is there a method to prove without the result for n∈Z ?" You did not address this at all. – fleablood Sep 27 '16 at 17:28
  • 2
    I haven't used the result for $n\in\mathbb{Z}$ though - I've taken $n=p/q$, and used standard algebraic techniques from there. – πr8 Sep 27 '16 at 17:29
  • You've converted $x^{p/q} = [x^{1/q}]^p$ and and relied on $p \in \mathbb Z$. – fleablood Sep 27 '16 at 17:35
  • 2
    @fleablood Right - I'm not sure how that's a problem. I might be missing something - my understanding of OP's question was that they wanted a way of proving what the derivative of $x^n$ was for non-integer, rational $n$, which didn't rely on knowing what the derivative of $x^n$ was for integer $n$. [I know OP says we're not assuming knowledge of differentiation, but it's the simplest way of referring to the limits in question] – πr8 Sep 27 '16 at 17:36
  • Hmmm, I intepretted the OPs comment differently.. namely that you prove it for $n \in Z$ and then apply $x^n = x^{p/q} = (x^{1/q})^p = y^{n'}$ with $n' \in Z$ and the OP (for reasons unknown) wanted a direct proof for rational $p/q$ without an intermediate integer step. – fleablood Sep 27 '16 at 18:55
  • @fleablood I'm still a little confused - are you taking issue with expressing a rational number as a ratio of integers? – πr8 Sep 27 '16 at 19:14
  • No. We prove this result for n where n is an integer directly. Then if r is rational we simply replace $z^r $ with $w^n $ where $r =n/m$ and $w = mth root (z)$ for integer n and m, and appeal to the result for integers. The op distinctly said he does not want to do that. He wants to prove it for rational numbers without first proving it for integers. I don't know why the op wants to do that, but that was has question. Your proof requires showing it for integers. – fleablood Sep 27 '16 at 19:54
3

In my opinion:

"Since Z⊂Q , why can't we prove this result only for n∈Q

?"

Is there a method to prove limx→a $x^{p/q}−a^{p/q}/x−a=p/q⋅a^{p/q−1}$ without the result for n∈Z ? "

I'd say "Not really".

We've only defined $x^{p/q}$ as ${\sqrt[q]{x}}^p$ and haven't really explored any ideas of what this can mean other than $x^r \approx y^n$ by "plugging in" $\sqrt[q]x$ for $y$ and $n = rq \in \mathbb Z$ for $n$.

So if we've only defined $x^r$ in terms of integers we can't do anything else.... yet.

Once we get some results about $x^r$ that don't rely on an integer definition we can. Maybe.

fleablood
  • 124,253
2

Proving $$\lim_{x\to a} \frac{x^{\frac{p}{q}}-a^{\frac{p}{q}}}{x-a}={\frac{p}{q}}\cdot a^{{\frac{p}{q}}-1}\,(p,q\in{\mathbb{Z}})$$

First,let $$y=x^{\frac{p}{q}}-a^{\frac{p}{q}} $$

Then,we can know $$y+a^{\frac{p}{q}}=x^{\frac{p}{q}}$$ $$(y+a^{\frac{p}{q}})^q=x^p$$ Use Binomial theorem: $$(y+a^{\frac{p}{q}})^q=y^q+q*y^{q-1}*a^{\frac{p}{q}}+...+q*y*a^\frac{p(q-1)}{q}+a^p$$ So,we can get:$$x^p-a^p=y^q+q*y^{q-1}*a^{\frac{p}{q}}+...+q*y*a^\frac{p(q-1)}{q}$$ For the identity: $$(x^p-a^p)=(x-a)(x^{p-1}+x^{p-2}*a+...+x*a^{p-2}+a^{p-1})$$ So $$x-a=\frac{y^q+q*y^{q-1}*a^{\frac{p}{q}}+...+q*y*a^\frac{p(q-1)}{q}}{x^{p-1}+x^{p-2}*a+...+x*a^{p-2}+a^{p-1}}$$ Finally, $$\frac{y}{x-a}=\frac{x^{p-1}+x^{p-2}*a+...+x*a^{p-2}+a^{p-1}}{y^{q-1}+q*y^{q-2}*a^{\frac{p}{q}}+...+q*a^\frac{p(q-1)}{q}}$$ Obviously:$$\lim_{x\to a}y=0 \,\,\,and\,\,\, \lim_{y\to 0}c*y^k=0({c}\in{\mathbb{R}},{k}\in{\mathbb{N^{+}}})$$ At last:$$\lim_{x\to a} \frac{x^{\frac{p}{q}}-a^{\frac{p}{q}}}{x-a}=\lim_{x\to a}\frac{y}{x-a}=\frac{p*a^{p-1}}{q*a^\frac{p(q-1)}{q}}=\frac{p}{q}*a^{p-1-\frac{p(q-1)}{q}}=\frac{p}{q}*a^{\frac{p}{q}-1}$$

Kudryavka
  • 21
1

A hint: according to Mean Value Theorem $\exists \epsilon \in (x,a):f(x)-f(a)=f'(\epsilon)(x-a)$ where $f(x)=x^{\frac{p}{q}}$ and $f'(x)$ is also continuous, assuming $a$ is not $0$. If $a=0$, skip MVT, apply direct substitution.

rtybase
  • 16,907
  • +1 for being the only answer so far that actually paid attention to the OPs actual question. – fleablood Sep 27 '16 at 17:32
  • Opps ... I just spotted the updated <<Please note that this question was asked by one of my students who doesn't know differentiation yet nor Lhopital nor mean value theorems>> well ... too late – rtybase Sep 27 '16 at 17:36
1

How's this? (Depending on what you mean by "only" the rationals, this may not be what you were looking for, but it doesn't invoke the result for integers.)

$$\lim_{x\to a}\frac{x^n-a^n}{x-a} = \lim_{x\to a}\left(\frac{a^n}{a^n}\cdot\frac{x^n-a^n}{\ln(x^n)-\ln(a^n)}\cdot\frac{\ln(x^n)-\ln(a^n)}{x-a}\cdot\frac{a}{a}\right) = \lim_{x\to a}\left(\frac{a^n(x^n-a^n)}{a^n\ln\left(\frac{x^n}{a^n}\right)}\cdot\frac{an\ln\left(\frac{x}{a}\right)}{a(x-a)}\right) = \lim_{x\to a}\left(\frac{a^n}{\ln\left(\frac{x^n-a^n}{a^n}+1\right)^\frac{a^n}{x^n-a^n}}\cdot\frac{n\ln\left(\frac{x-a}{a}+1\right)^\frac{a}{x-a}}{a}\right) = \frac{a^n}{\ln e}\cdot\frac{n\ln e}{a} = n\cdot a^{n-1}$$

user361424
  • 1,719
0

$\dfrac{x^n-a^n}{x-a}\;$ is the rate of variation at $x=a$ of the function $x^n$, hence the limit is, by definition, the derivative of $x^n$ at $x=a$, i.e. $$\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}.$$

Bernard
  • 175,478
  • 1
    This is somewhat circular: the derivative is defined to be the result of the limit, so in order to know what the derivative of $x^n$ is at $x=a$, you need to know what the limit is! – Clive Newstead Sep 27 '16 at 19:36
  • @Clive Newstead Oh! I see. I misunderstood what the problem was. – Bernard Sep 27 '16 at 19:47
0

One approach is to use the inequalities and squeeze theorem, but proving the desired inequalities is not so easy. I have already provided details of this approach in this answer and I advise you to have a look. However you will see that proving the desired inequalities starts with the integers and then from it we deduce the inequalities for the rational numbers.

You have to understand that the idea of rationals cannot be had without the idea of integers (you may perhaps use some crafty definition which avoids the term integers like "smallest field of characteristic $0$" but you can not escape the fact that the field of rationals is the field of quotients of integers).

Community
  • 1
Paramanand Singh
  • 87,309