Question about the proof of $\lim_{x\to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}$

Question

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:

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

Answer 1

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 $1$.

Answer 2

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 $.