Question about the proof of $\lim_{x\to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}$ Proof:
Let, $x-a=h\implies x=a+h$
$$\lim_{x\to a}\frac{x^{n}-a^{n}}{x-a}$$
$$\lim_{h\to 0}\frac{(a+h)^{n}-a^{n}}{h}$$
$$\lim_{h\to 0}\frac{a^n(1+\frac{h}{a})^{n}-a^{n}}{h}$$
Since $h\to0$, $h$ can be supposed to be less than $a$. So, $|\frac{h}{a}|<1$. Therefore, $(1+\frac{h}{a})^n$ can be expanded with the help of the binomial theorem.
$$\text{rest of the proof...}$$
Questions:

*

*Why do we need to assume that $h$ is smaller than $a$? Does this property not hold if $h$ is greater than or equal to $a$?

*Why does $|\frac{h}{a}|<1$ need to be true for us to expand $(1+\frac{h}{a})^n$ using the binomial theorem?

 A: You problem lies in "$n$".
Writing the limit theorem precisely:
For any positive integer $n$, $$\lim_{x\to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1} \tag{A}\label{A}$$ where $a$ is any non-zero real number.
Now, for proving this theorem after this step: $$\lim_{h\to 0}\frac{a^n(1+\frac{h}{a})^{n}-a^{n}}{h}$$
and after the expansion of $(1+\frac{h}{a})^n$ you are correct to say that "higher-order terms become zero anyways (after inputting h=0). So, why does $|\frac{h}{a}|$ need to be less than 1?"
Indeed $|\frac{h}{a}|<1$ is not necessary for this proof.
Also $|\frac{h}{a}|<1$ does not need to be true for us to expand $(1+\frac{h}{a})^n$ using the Binomial theorem when $n$ is positive integer
But perhaps you are trying to show that $(\ref{A})$ holds even when $n$ is a rational number and $a$ is positive.
Now, the binomial expansion, $$(1+x)^n= 1+ nx + \frac{n(n-1)}{ 2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3+\cdots \tag{B}\label{B}$$ is true for all rational number $n$ (and has infinite no. of terms when $n \in \mathbb{Q} \setminus \mathbb{W}$) only when $|x|<1$ and certainly in this case, $|\frac{h}{a}|<1$ is true, so yes $(\ref{A})$ holds even when $n$ is rational.
Note that, the binomial expansion $(\ref{B})$ is true when $n$ is a positive integer (and has exactly $n+1$ terms) regardless of $|x|$ being less than or greater than or equal to $1$.
A: Another way.
Use
$x^n-a^n
=(x-a)\sum_{k=0}^{n-1}x^ka^{n-1-k}
$.
Then,
if
$0 < a-c < x < a+c$,
$\begin{array}\\
\dfrac{x^n-a^n}{x-a}
&=\sum_{k=0}^{n-1}x^ka^{n-1-k}\\
\text{so}\\ 
\dfrac{x^n-a^n}{x-a}
&\lt\sum_{k=0}^{n-1}(a+c)^ka^{n-1-k}\\
&\lt \sum_{k=0}^{n-1}(a+c)^{n-1}\\
&=n(a+c)^{n-1}\\
\text{and}\\
\dfrac{x^n-a^n}{x-a}
&\gt\sum_{k=0}^{n-1}(a-c)^ka^{n-1-k}\\
&\gt \sum_{k=0}^{n-1}(a-c)^{n-1}\\
&=n(a-c)^{n-1}\\
\end{array}
$
I'll work with
the first inequality here.
The second is similar.
If $0 < z < 1$
then
$(1+z)^m
=1+\sum_{j=1}^m \binom{m}{j}z^j
\le 1+\sum_{j=1}^m \binom{m}{j}z
=1+(2^m-1)z
$
so
$(1+z)^m-1
\le (2^m-1)z
$.
Therefore
$\begin{array}\\
\dfrac{x^n-a^n}{x-a}-na^{n-1}
&\lt n(a+c)^{n-1}-na^{n-1}\\
&=na^{n-1}((1+c/a)^{n-1}-1)\\
\text{so if } c<a\\
\dfrac{x^n-a^n}{x-a}-na^{n-1}
&\lt na^{n-1}((1+c/a)^{n-1}-1)\\
&\lt na^{n-1}(2^{n-1}-1)(c/a)\\
\end{array}
$
so if
$c < \dfrac{a\epsilon}{na^{n-1}(2^{n-1}-1)}
$
(remembering that
$x < a+c$)
then
$\dfrac{x^n-a^n}{x-a}-na^{n-1}
\lt \epsilon
$.
