Proving $\lim \frac{x^n-a^n}{x-a}=n\cdot a^{n-1}$ 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 $  ?
 A: 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 $\frac{y^n-1}{y-1}=y^{n-1}+y^{n-2}+\cdots+y+1$.
A: 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.
A: 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:$$\lim_{x\to a} \frac{x^{\frac{p}{q}}-a^{\frac{p}{q}}}{x-a}=\lim_{x\to a}\frac{y}{x-a}=\frac{p*a^{p-1}}{q*a^\frac{p(q-1)}{q}}=\frac{p}{q}*a^{p-1-\frac{p(q-1)}{q}}=\frac{p}{q}*a^{\frac{p}{q}-1}$$
A: 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.
A: 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.)
$$\lim_{x\to a}\frac{x^n-a^n}{x-a} = \lim_{x\to a}\left(\frac{a^n}{a^n}\cdot\frac{x^n-a^n}{\ln(x^n)-\ln(a^n)}\cdot\frac{\ln(x^n)-\ln(a^n)}{x-a}\cdot\frac{a}{a}\right) = \lim_{x\to a}\left(\frac{a^n(x^n-a^n)}{a^n\ln\left(\frac{x^n}{a^n}\right)}\cdot\frac{an\ln\left(\frac{x}{a}\right)}{a(x-a)}\right) = \lim_{x\to a}\left(\frac{a^n}{\ln\left(\frac{x^n-a^n}{a^n}+1\right)^\frac{a^n}{x^n-a^n}}\cdot\frac{n\ln\left(\frac{x-a}{a}+1\right)^\frac{a}{x-a}}{a}\right) = \frac{a^n}{\ln e}\cdot\frac{n\ln e}{a} = n\cdot a^{n-1}$$
A: $\dfrac{x^n-a^n}{x-a}\;$ 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.
$$\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}.$$
A: 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).
