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:

  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?
tryingtobeastoic
  • 3,176
  • 1
    The assumption $|h|<|a|$ isn't necessary for the proof if $n$ is a positive interger. I guess the writer mixed up things with series (non-interger or negative $n$). – A.Γ. Oct 01 '21 at 13:39
  • @YvesDaoust the higher-order terms become zero anyways (after inputting h=0). So, why does $|\frac{h}{a}|$ need to be less than 1? – tryingtobeastoic Oct 01 '21 at 13:42
  • To make $\left|\dfrac ha\right|^{n+1}\ll\left|\dfrac ha\right|^n$. Negligible. –  Oct 01 '21 at 13:45
  • 1
    The binomial theorem doesn't need any assumption on how big or small are the terms. If the presented proof was taken from a book, then I agree with @A.Γ. that it seems to be a sort of mix with another topics. – FormerMath Oct 01 '21 at 13:48
  • Related – tryingtobeastoic Oct 09 '21 at 10:52

2 Answers2

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 $1$.

Aman Kushwaha
  • 1,664
  • 1
    $\mathbb{W}=\mathbb{Z}^+ \cup {0}$ is the set of whole numbers. – Aman Kushwaha Oct 01 '21 at 14:23
  • 1
    Thanks for your great answer!! Extra question: can we apply the above formula to this: $$\lim_{x\to -2} \frac{x^2-(-2)^2}{x+2}$$? I get the correct answer by using the above formula, but you said $a$ has to be a positive real number, so I'm confused. – tryingtobeastoic Oct 01 '21 at 14:57
  • 1
    @tryingtobeastoic Yes, when $n$ is a positive integer, $a$ can be any real number. You can see in the proof $a$ being negative doesn't make any difference when $n$ is positive integer but $a>0$ is required when $n$ is any rational number other than whole number. – Aman Kushwaha Oct 01 '21 at 15:07
  • 1
    Yeah, I wrote $a$ is positive real number everywhere because I was trying to limit my focus to $n$ but now I have edited my answer to make it more generalised so as to cover your extra question. I hope you are not confused now and you can see that the answer without edit had no problem either, it's just that I didn't include the case when a is negative. – Aman Kushwaha Oct 01 '21 at 15:13
  • 1
    Also, $a=0$ is fine unless $n \neq 1$. So, for any positive integer $n$, $a$ should be non-zero – Aman Kushwaha Oct 01 '21 at 15:21
  • So, for any positive integer or rational number, $a$ has to be non-zero. The only exception is when $n=1$, right? – tryingtobeastoic Oct 08 '21 at 11:14
  • 1
    @tryingtobeastoic $a$ has to be non-zero when $n=1$ because when $n=1$, RHS of (A), i.e., $a^0$ would be undefined at $a=0$ however the limit exists and is equal to 1. When $n \neq1 ; (n \in \Bbb Z^+)$, $a$ can be any real number including zero. You can verify that the limit will be zero and so will be the RHS for $a=0$ when $n \in {2,3,4,\ldots}$. – Aman Kushwaha Oct 08 '21 at 16:48
  • 1
    Coming to the case when $n$ is any rational number, $a$ has to be non-zero again here. You can check what happens at $n=0$ or $n=0.5$ or any negative $n$ in (A) if $a=0$. Therefore better to have non-zero $a$ always (also because the special case of $a=0$ trivial in itself that you don't have to use this formula for calculating the limit.) – Aman Kushwaha Oct 08 '21 at 16:57
  • Thanks for your detailed comments!!!! I had a question: you earlier said that if $n$ is a positive integer, then $a$ can be any real number (including zero). However, if $a = 0$, then $\frac{h}{a}=\frac{h}{0}$ becomes undefined; doesn't our proof become invalid then? – tryingtobeastoic Oct 09 '21 at 04:25
  • 1
    @tryingtobeastoic Man, What I said earlier was "if $n$ is a positive integer and $n\neq 1$ then $a$ can be any real number including zero in (A) ". I never said the same proof will work for this trivial case of $a=0$. Isn't it obvious that when $a=0$, $ \small \lim\limits_{x\to a}\frac{x^{n}-a^{n}}{x-a}=\lim\limits_{x\to 0}\frac{x^{n}}{x}=\lim\limits_{x\to 0} x^{n-1} =0 ;(\because n\in {2,3,4, \ldots})$ which is equal to the RHS of (A), i.e., $n 0^{n-1}=0$. Now why would you go for a binomial proof?And how would $a+h$ be equal to $a(1+\frac{h}{a})$ when $a=0$. – Aman Kushwaha Oct 09 '21 at 07:34
0

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 $.

marty cohen
  • 107,799