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 $  ?

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 $\frac{y^n-1}{y-1}=y^{n-1}+y^{n-2}+\cdots+y+1$.
