Understanding the $\varepsilon-\delta$ proof of $\underset{x\to 3}{\lim}{f(x)}=9$ with $f(x)=x^2$

Question

This is the proof I have:

Choose some $\epsilon > $0 then we can show that $\exists \delta_\epsilon$ s.t. $|x^2-9| < \epsilon$ $\forall |x-3| < \delta_\epsilon$. $|x^2-9|=|x-3||x+3| \leq |x-3||x-3+6| \leq |x-3|(|x-3|+6)$. If $|x-3|< \delta_\epsilon$ then $|x^2-9|\leq \delta_\epsilon(\delta_\epsilon+6) <\epsilon$. Set $\delta_\epsilon = min \{1, \frac{\epsilon}{7}\}$ then we have $\forall \epsilon > 0$ $\exists \delta_\epsilon>0$ s.t. $|x^2-9| < \epsilon$ $\forall |x-3| < \delta_\epsilon$.

The second I didn't understand was why we set $\delta_\epsilon = min \{1, \frac{\epsilon}{7}\}$. Did someone just experiment with different settings of $\delta_\epsilon$ and find that that was the that worked or is there some intuition behind it?

Answer 1

Let $k>0$ be a constant. The idea is to suppose that $\delta <k$, then you can find an upper bound for the factor $\delta+6$ will satisfy:

$$\delta+6 <k+6$$

and finally :

$$ \delta(\delta+6)<\delta(6+k)\leq \varepsilon, $$

this tells us that

$$\delta(6+k)\leq \varepsilon$$

and this implies $$\displaystyle{\delta\leq\frac{\varepsilon}{6+k}}$$

then to have both $\displaystyle{\delta\leq\frac{\varepsilon}{6+k}}$ and $\delta <k$ you can say let $\delta$ be the minimum between $\frac{\varepsilon}{6+k}$ and $k$.

Answer 2

$\delta(\delta+6) <\epsilon \iff \delta^2+6\delta<\epsilon \iff (\delta+3)^2<\epsilon+9 \iff \delta<\sqrt{\epsilon+9}-3$.

So any $\delta \in (0,\sqrt{\epsilon+9}-3)$ works. Observe that $\delta_\epsilon \in (0,\sqrt{\epsilon+9}-3]$.

Answer 3

This is intended as a detailed explanation of why this type of argument works.

Suppose $f$ is an algebraic function in a single variable $x$ and $L\in\mathbb{R}$.

Suppose you wish to prove by the $\epsilon$-$\delta$ method that

$$ \lim_{x\to a}f(x)=L$$

If it happens to be the case that within some radius $r$ of $a$, the quantity

$$\left\vert \frac{f(x)-L}{x-a}\right\lvert$$

has an upper bound $B$ when $0<\vert x-a\vert<r$

then let $\delta=\min\left\{r,\dfrac{\epsilon}{B}\right\}$.

In this example, one wishes to show that

$$\lim_{x\to3}x^2=9$$

So we would like to first try to find an upper bound on the value of

$$\left\vert\frac{x^2-9}{x-3} \right\vert$$

for some $r$ [Hint: $r=1$ often works] when $0<\vert x-3\vert<r$.

Now since

$$ \left\vert\frac{x^2-9}{x-3} \right\vert=\vert x+3\vert $$

we must find if $\vert x+3\vert$ has an upper bound for some interval about $3$.

Certainly this is a case where we may use $r=1$ because for $0<\vert x-3\vert<1$ it follows that $2<x<4$ therefore $5<x+3<7$. Thus $7$ is an upper bound on

$$ \vert x+3\vert = \left\vert\frac{x^2-9}{x-3} \right\vert<7\text{ when } 0<\vert x-3\vert<1$$

So we let $\delta=\min\left\{1,\frac{\epsilon}{7}\right\}$

Now we may begin with the proof, since all the preceding occurs "behind the scenes."

PROOF:

Let $\epsilon>0$.

Let $\delta=\min\left\{1,\frac{\epsilon}{7}\right\}$

Then if $\vert x-3\vert<\delta$ it follows that $\vert x-3\vert<1$ and $\vert x-3\vert<\frac{\epsilon}{7}$. So $\vert x+3\vert<7$.

But since $\vert x+3\vert=\left\vert\frac{x^2-9}{x-3} \right\vert$ we have both that

\begin{eqnarray} \left\vert\frac{x^2-9}{x-3} \right\vert&<&7\text{ and }\\ \vert x-3\vert&<&\frac{\epsilon}{7} \end{eqnarray}

Therefore

$$ \vert x^2-9\vert=\left\vert\frac{x^2-9}{x-3} \right\vert\cdot\vert x-3\vert<7\cdot\frac{\epsilon}{7}=\epsilon $$

Which is the end of the proof.