Proving $\lim\limits_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}$ by induction

Question

Prove by induction that $$\lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}.$$

I did a strange proof using two initial results: We know that result is true for $n=1$ and $n=2$. Assuming the result is true for $n=k-1$ and $n=k$, I can prove the result for $n=k+1$. For this I used my assumption for the case of $n=k$, and used a random multiplier to get the desired result: $$\lim_{x \to a} \frac{x^{k}-a^{k}}{x-a}·\frac{x+a}{x+a}.$$ During the simplification I had to use my $n=k-1$ case result as well.

Is this proof OK in terms of induction principle? I can show my whole working if needed.

Thank you.

Edit

whole proof:

$\lim_{x \to a} \frac{x^{k}-a^{k}}{x-a}·\frac{x+a}{x+a}=k.a^{k-1}$ (I'll omit limit notation for clarity)

$$\frac{x^{k}-a^{k}}{x-a}·\frac{x+a}{x+a}=k.a^{k-1}$$

$$\frac{1}{2a}\frac{x^{k+1}+ax^k-a^kx-a^{k+1}}{x-a}=k.a^{k-1}$$

$$\frac{1}{2a}(\frac{x^{k+1}-a^{k+1}}{x-a}+\frac{ax(x^{k-1}-a^{k-1})}{x-a})=k.a^{k-1}$$

$$\frac{1}{2a}[\frac{x^{k+1}-a^{k+1}}{x-a}+{a^2(k-1).a^{k-2}}]=k.a^{k-1}$$

Hence, $$\lim_{x \to a} \frac{x^{k+1}-a^{k+1}}{x-a}=(k+1)a^{k}.$$

Answer 1

Hint $$\frac{x^{n+1}-a^{n+1}}{x-a}=\frac{x^{n+1}-ax^n}{x-a}+\frac{ax^n-a^{n+1}}{x-a}$$

Answer 2

$$x^{n+1}-a^{n+1}=(x-a)(x^n+a^n)+ax(x^{n-1}-a^{n-1})$$

Answer 3

My algebra would be:

Suppose $\lim_\limits{x\to a} \frac {x^n-a^n}{x-a} = na^{n-1}$

Show that $\lim_\limits{x\to a} \frac {x^{n+1}-a^{n+1}}{x-a} = na^{n}$

$\frac {x\cdot x^{n} - a\cdot a^{n}}{x-a}$

Add and subtract the same term such that your expression will factor into something that lest you use the inductive hypothesis.

$\frac {x\cdot x^{n} -xa^n+xa^n- a\cdot a^{n}}{x-a}\\ \frac {x(x^{n} -a^n)}{x-a} +\frac{x -a}{x-a}a^n$

Then invoke the inductive hypothesis. Evaluate the limits for what remains and you are done.