$|x-2|$ as a factor of $|x^n-2^n|$ as a limit of function

Question

I haven't posted in a while but I do have a question on factoring a specific term out of a particular polynomial function and I'm stuck at some point in the process. Please, let's see so that my nightmares go away.

Say we seek to prove that $\:\lim_{x\to2}\large\:\frac{|x^n-2^n|}{x-2}$$\:=n2^{n-1}$

Thus, $\large\:|\frac{x^n-2^n}{x-2}-$$2n^{n-1}\large|$$\:=\large|\frac{x^n-2^n-(x-2)2n^{n-1}}{x-2}|$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\large|$$(x-2)\cdot\large\frac{x^{n-1}+x^{n-2}\cdot2+x^{n-3}\cdot2^2+.\:.\:.\:+x^2\cdot2^{n-3}+x\cdot2^{n-2}+2^{n-1}-n2^{n-1}}{x-2}|$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\small|x^{n-1}+x^{n-2}\cdot2+x^{n-3}\cdot2^2+.\:.\:.\:+x^2\cdot2^{n-3}+x\cdot2^{n-2}+2^{n-1}-n2^{n-1}|$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\small|x^{n-1}+x^{n-2}\cdot2+x^{n-3}\cdot2^2+.\:.\:.\:+x^2\cdot2^{n-3}+x\cdot2^{n-2}+(1-n)(2^{n-1})|$

Now, since we know $\:|x-2|\:$is a factor of$\:|x^n-2^n|\:$the big question becomes how could we use such a factor on :

$\small|x^{n-1}+x^{n-2}\cdot2+x^{n-3}\cdot2^2+.\:.\:.\:+x^2\cdot2^{n-3}+x\cdot2^{n-2}+(1-n)(2^{n-1})|\:$?

So that, we would use a simple restriction on$\:|x-2|<1\:$and then have a minimized$\:\delta(\epsilon)\:=\imath\eta f\{1,\frac{\large\epsilon}{\phi}\}\:$where we let$\:\phi :=\:$the multiple that is strictly larger than the multiple of $\:|x-2|\:$that we're looking for.

Thanks!

Answer 1

For $a≠0$:

\begin{align} \dfrac{x^n-a^n}{x-a} - na^{n-1} &= \sum_{k=0}^{n-1} (x^ka^{n-k-1}-a^{n-1})\\ &= \sum_{k=1}^{n-1} a^{n-k-1} (x^k-a^{k})\\ \end{align}

Triangle inequality says:

$$\left|\dfrac{x^n-a^n}{x-a} - na^{n-1}\right| \le \sum_{k=1}^{n-1} |a^{n-k-1}||x^k-a^k|$$

If $|x-a|\le |a|$, then \begin{align} |x^k-a^k|&\le|x-a|\sum_{j=0}^{k-1}|x|^j|a|^{k-j-1}\\ &\le |x-a||a|^{k-1}(2^k-1) \end{align}

Finally, we have:

\begin{align}\left|\dfrac{x^n-a^n}{x-a} - na^{n-1}\right|& \le \sum_{k=1}^{n-1} |a^{n-k-1}||x^k-a^k|\\ &\le |x-a||a|^{n-2}\sum_{k=1}^{n-1} (2^k-1)\\ &\le |x-a||a|^{n-2}(2^n-n) \end{align}

Let $\delta = \min(|a|,\dfrac{\epsilon}{|a|^{n-2}(2^n-n)})$

Answer 2

Note: The proof has been completely rewritten.

More generally, you want to show that, if $a > 0$ then $\lim_{x\to a}\frac{x^n-a^n}{x-a} =na^{n-1} $.

You have the right start:

$\frac{x^n-a^n}{x-a} =\sum_{k=0}^{n-1} x^k a^{n-1-k} $.

What you need to show is that, for $x$ close to $a$, $\sum_{k=0}^{n-1} x^k a^{n-1-k}$ is close to $n a^{n-1}$.

What makes this reasonable is that, for $x$ close to $a$, $x^k a^{n-1-k}$ is close to $a^{n-1}$ so the sum of these is close to $na^{n-1}$.

To make this rigorous, suppose $|x-a| < c$, where, eventually, we will make $c$ as small as we need. We certainly need $c < a$.

Then $-c < x-a < c$ or $a-c < x < a+c$.

First, we get an upper bound on the terms.

$\begin{array}\\ x^ka^{n-1-k} &\lt (a+c)^ka^{n-1-k}\\ &= a^k(1+c/a)^ka^{n-1-k}\\ &= (1+c/a)^ka^{n-1}\\ \end{array} $

We now use this lemma:

If $u > 0, v > 0,$ and $uv < 1$ then $(1+u)^v < \frac1{1-uv} $. If $uv < \frac12$, then $(1+u)^v < 1+2uv $.

Proof:

If $uv \le \frac12$ then

$\begin{array}\\ (1+u)^v &= e^{v\ln(1+u)}\\ &\le e^{vu} \qquad\text{(since }\ln(1+u) < u)\\ &\le \frac1{1-uv} \qquad\text{(since }e^x \le \frac1{1-x} \text{ for } 0 \le x < 1)\\ &\le 1+2uv \quad\text{(since } \frac1{1-x}=1+\frac{x}{1-x}\le 1+2x)\\ \end{array} $

Therefore, if $\frac{cn}{a} \le \frac12 $, then $(1+c/a)^k \lt 1+\frac{2ck}{a} $ so that $x^ka^{n-1-k} \lt (1+\frac{2ck}{a})a^{n-1} $.

Summing,

$\begin{array}\\ \sum_{k=0}^{n-1} x^ka^{n-1-k} &\lt \sum_{k=0}^{n-1} (1+\frac{2ck}{a})a^{n-1}\\ &= na^{n-1}+2ca^{n-2}\frac{n(n-1)}{2}\\ &< na^{n-1}+cn^2a^{n-2}\\ \end{array} $

Therefore, if $cn^2a^{n-2} < \epsilon$, $\sum_{k=0}^{n-1} x^ka^{n-1-k} < na^{n-1}+\epsilon $.

For a lower bound,

$\begin{array}\\ x^ka^{n-1-k} &\ge (a-c)^ka^{n-1-k}\\ &= a^k(1-c/a)^ka^{n-1-k}\\ &= (1-c/a)^ka^{n-1}\\ \end{array} $

We now use this lemma:

If $u > 0, v > 0,$ $uv < 1$, and $\frac{2uv}{1-u} < \frac12$, then $(1-u)^v \gt 1-\frac{2uv}{1-u} $.

Proof.

$\begin{array}\\ (1-u)^{-v} &=(\frac1{1-u})^{v}\\ &=(1+\frac{u}{1-u})^{v}\\ &\le 1+\frac{2uv}{1-u} \qquad\text{(by the lemma above)}\\ &= \frac{1-u+2uv}{1-u}\\ \text{so}\\ (1-u)^{v} &\ge \frac{1-u}{1-u+2uv}\\ &= 1-\frac{2uv}{1-u+2uv}\\ &\gt 1-\frac{2uv}{1-u}\\ \end{array} $

Therefore, if $\frac{2cn}{a(1-c/a)} \le \frac12 $, or $\frac{2cn}{a-c} \le \frac12 $,

then $(1-c/a)^k \gt 1-\frac{2ck}{a(1-c/a)} = 1-\frac{2ck}{a-c} $ so that $x^ka^{n-1-k} \gt (1-\frac{2ck}{a-c})a^{n-1} $.

Summing, assuming in addition that $a-c > \frac{a}{2}$ or $c < \frac{a}{2} $,

$\begin{array}\\ \sum_{k=0}^{n-1} x^ka^{n-1-k} &\gt \sum_{k=0}^{n-1} (1-\frac{2ck}{a-c})a^{n-1}\\ &= na^{n-1}-\frac{2ca^{n-1}}{a-c}\frac{n(n-1)}{2}\\ &\gt na^{n-1}-\frac{cn^2a^{n-1}}{a-c}\\ &\gt na^{n-1}-2cn^2a^{n-2}\\ \end{array} $

Therefore, if $2cn^2a^{n-2} < \epsilon$ and $c < \frac{a}{2} $, $\sum_{k=0}^{n-1} x^ka^{n-1-k} \gt na^{n-1}-\epsilon $.