Epsilon-delta proof for $\lim_{x \rightarrow a} \frac{x^n - a^n}{x -a}$ I'm having trouble finding the right way to approach this limit.
$$ \left| \frac{x^n - a^n}{x-a} - na^{n-1} \right| < \varepsilon, \text{ given } |x-a| < \delta $$
I've tried rewriting $\frac{x^n - a^n}{x-a}$ as $\sum_{k=0}^n x^ka^{n-k}$, but that made it hard to reintroduce a $\delta$ into the inequality. I've also tried assuming $\delta < 1$ and rewritting the numerator as $(a+1)^n - a^n$, but similarly ran into problems with a disappearing delta.
Can I please have a nudge in the right direction?
 A: The sum should be $\sum_{k=0}^{n-1} x^k a^{n-k-1}$. Note that plugging in $x=a$ immediately gives you $na^{n-1}$, so appealing to continuity of $x \mapsto \sum_{k=0}^{n-1} x^k a^{n-k-1}$ gives you the desired limit without any $\epsilon-\delta$.

However, if you need to still do that argument, try
$$\left|\sum_{k=0}^{n-1} x^k a^{n-k-1} - na^{n-1}\right|
\le \sum_{k=0}^{n-1} |a|^{n-k-1} |x^k - a^k|.\tag{$*$}$$
The terms $|x^k - a^k|$ can be bounded using $|x-a| < \delta$ as follows.
$$|x^k-a^k| \le |x-a|\sum_{j=0}^{k-1} |x^j a^{k-j-1}|.$$
If you assume $\delta < 1$ you can use $|x| \le |a|+1$ to bound the terms $|x^j a^{k-j-1}|$ by some constant involving $|a|$, so you end up with something like $|x^k - a^k| \le C_k \delta$. Plugging this back into the earlier sum ($*$) and doing some more accounting will give you an overall bound of $C\delta$ for some constant $C$.
A: Hint: This is much easier if you rewrite your limit to
$$
\lim_{h\to 0}\frac{(a+h)^n-a^n}h
$$
using the substitution $x\mapsto a+h$. And then use the binomial theorem to expand the numerator.
You can do this without substitution too, by rewriting $x^n=(a-(x-a))^n$ and expanding that (but keeping every $(x-a)^k$ unexpanded). It's the same thing, possibly more conceptually transparent, but a lot more writing.
A: Your expression for  $\frac{x^n - a^n}{x-a}$ is $\sum_{k=0}^{n-1} x^ka^{n-1-k}$. So  you  might want to consider $\sum_{k=0}^{n-1} x^k a^{n-1-k}-na^{n-1}=\sum_{k=0}^{n-1}(x^k-a^k)a^{n-k-1}$.
A: My 2 cents. If can be accepted $f(x)=x^k$ continuity and having fixed $n$:
$$\left| \frac{x^n - a^n}{x-a} - na^{n-1} \right| = \left| \frac{(x - a)(x^{n-1}+ax^{n-2}+\cdots + a^{n-1})}{x-a} - na^{n-1} \right| \leqslant \\
|x^{n-1}-a^{n-1}| + |ax^{n-2} - a^{n-1}|+\cdots + |x a^{n-2}-a^{n-1}|+|a^{n-1}-a^{n-1}| \leqslant (n-1)\cdot\frac{\varepsilon}{n-1}\\
$$
A: After simplification,
$$\left|\frac{(a+\delta)^n-a^n}{a+\delta-a}\right|=\left|\binom n1a^{n-1}+\binom n2a^{n-2}\delta+\cdots\binom nn\delta^{n-1}\right|
\\=na^{n-1}\left|1+2(n-1)\rho+\cdots(n-1)\binom{n-1}{n-1}\rho^{n-1}\right|
\\\le na^{n-1}|1+m\rho|$$ where $\rho:=\dfrac{\delta}{a}<1$ and $m$ only depends on $n$. Hence
$$\delta=a\min\left(\frac\epsilon m,1\right)$$ does the trick.
