Delta Epsilon Proof: An Overview

[karush]

View attachment 2404

these proofs are always confusing but here's my take on it..

since $x\rightarrow +\infty$ we don't need absolute values and since
$
\displaystyle
\frac{1}{10^2}=0.01
$
then we could use $N=10$ letting $L=0$ since it is a horz asymptote then we have

$
\displaystyle
\left[\frac{1}{x^2}-0\right]<0.01
$
by observation if $x>10$ this would be true.
also, if $\delta$ is the distance between $N$ and $x$

sure this isn't the full story.

 

[Prove It]

To prove that $\displaystyle \begin{align*} \lim_{x \to \infty} f(x) = L \end{align*}$, you have to show that for all $\displaystyle \begin{align*} \epsilon > 0 \end{align*}$ there exists an N such that $\displaystyle \begin{align*} x > N \implies \left| f(x) - L \right| < \epsilon \end{align*}$.

So in your case, to show $\displaystyle \begin{align*} \lim_{x \to \infty} \frac{1}{x^2} = 0 \end{align*}$, you need to show that $\displaystyle \begin{align*} x > N \implies \left| \frac{1}{x^2} - 0 \right| < \epsilon \end{align*}$.

So the scratch work to find a suitable N:

$\displaystyle \begin{align*} \left| \frac{1}{x^2} - 0 \right| &< \epsilon \\ \left| \frac{1}{x^2} \right| &< \epsilon \\ \frac{1}{x^2} &< \epsilon \\ 1 &< \epsilon \, x^2 \\ \frac{1}{\epsilon} &< x^2 \\ x^2 &> \frac{1}{\epsilon} \\ |x| &> \frac{1}{\sqrt{\epsilon}} \end{align*}$

and since we want $\displaystyle \begin{align*} x \to \infty \end{align*}$ we can assume x is some big positive number, giving $\displaystyle \begin{align*} x > \frac{1}{\sqrt{\epsilon}} \end{align*}$. So that means we can set $\displaystyle \begin{align*} N = \frac{1}{\sqrt{\epsilon}} \end{align*}$.

The actual proof will be to start with setting $\displaystyle \begin{align*} N = \frac{1}{\sqrt{\epsilon}} \end{align*}$ and using that to show that if $\displaystyle \begin{align*} x > \frac{1}{\sqrt{\epsilon}} \end{align*}$ then $\displaystyle \begin{align*} \frac{1}{x^2} < \epsilon \end{align*}$ (which should be pretty easy considering that every step in the scratch work is reversible)...