How Is $\delta = \sqrt{9+\epsilon}-3$ the Largest Choice in a Limit Proof?

[Dethrone]

Verify, by a geometric argument, that the largest possible choice of $\delta$ for showing that $\lim_{{x}\to{3}}x^2=9$ is $\delta = \sqrt{9+\epsilon}-3$

I have no clue, hints?

 

[MarkFL]

With $f(x)=x^2$, I would begin with:

\(\displaystyle f(3)+\varepsilon=f(3+\delta)\)

 

[Dethrone]

I can expand it...
$f(3+\delta)=9+6\delta +(\delta)^2$, but I still don't see a geometric argument from that :(

 

[MarkFL]

Draw the curve $f(x)=x^2$. Now, on the $x$-axis at $x=3$, draw a vertical line segment up to the curve. Where this line touches the curve, draw a horizontal line segment to the left, until it reaches the $y$-axis...this is $f(3)$. Now, go back to the vertical line, and choose a distance $\delta$ from $x=3$, which will be $3+\delta$. Draw a line up to the curve, and the to the left to the $y$-axis. It will be at $f(3)+\varepsilon$. Thus we see from this construction, that we must have:

\(\displaystyle f(3)+\varepsilon=f(3+\delta)\)

\(\displaystyle 9+\varepsilon=(3+\delta)^2\)

Instead of expanding, use the square root property, and take the positive root, and you will have shown what is required. :D