Finding Maximum Delta for a Limit Involving a Quadratic Function

[ver_mathstats]

Homework Statement

Find a maximum value of δ.

Relevant Equations

limx→3 (x^2) =9.

Consider lim_x→3x^2=9.
Find a maximum value of δ such that:
|x² - 9|<0.009 if |x-3|<δ

I just learned how to do this today and I am quite comfortable doing this if the function is linear, however now I am struggling with working with quadratic functions.

So far this is what I have come up with:

-0.009<x²-9<0.009

-0.009<(x-3)(x+3)<0.009

This is where I begin to get confused. Should I first solve it using epsilon and delta? And then use 0.009?

 

[fresh_42]

Draw a picture! Of course not with a true scale, exaggerate. You could solve $x^2-9 = 0.009$ and observe how far away from $3$ your values are allowed to be.

 

[Delta2]

It all boils down to solving a system of inequalities that involve $\delta$ and $\epsilon$.

To formulate the system I would start with $$|x-3|<\delta \Rightarrow -\delta<x-3<\delta \Rightarrow 6-\delta<x+3<6+\delta$$

Now by combining the last two inequalities what inequalities can you infer for the product $x^2-9=(x-3)(x+3)$ which has to satisfy $|x^2-9|<\epsilon$. You got to be careful in the combining though , cause one can't always multiply inequalities and get an inequality with the same direction.