Originally Posted by

**p0oint** Hello!

I know how to solve this task but I have certain things that I do not understand.

Solution.

$\displaystyle [|x^2-9|< \epsilon $ if $\displaystyle 0 < |x-3| < \delta$

Because |x-3| occurs on the right side of this "if statement", it will be helpful to factor the left side to introduce a factor of |x-3|.

$\displaystyle |x+3||x-3|< \epsilon$ if $\displaystyle 0<|x-3|<\delta$

Using the triangle inequality:

$\displaystyle |x+3|=|(x-3)+6| \leq |x-3| + 6$

By multiplying the statement above with |x-3| (which is positive) we got:

$\displaystyle |x+3||x-3| \leq (|x-3| +6)|x-3| $

Now clearly $\displaystyle 6+|x-3|< 6 + \delta$ and $\displaystyle |x-3|<\delta$

By multiplying both inequalities (I am not sure if this step is valid, please confirm me) I got:

$\displaystyle (6+|x-3|)|x-3|< (6 + \delta)\delta$

We can conclude that:

$\displaystyle |x+3||x-3| < (6 + \delta)\delta$ if $\displaystyle 0 < |x-3| < \delta$

Now isn't $\displaystyle \epsilon = (6 + \delta)\delta$ ??

Why in my book they choose $\displaystyle (6 + \delta)\delta \leq \epsilon$ ?

And why they choose $\displaystyle \delta \geq 1 $ and with that restriction

$\displaystyle (6 + \delta)\delta \leq 7\delta$ and $\displaystyle 7\delta \leq \epsilon$.

Finally I don't understand why $\displaystyle \delta = min(e/7,1)$.

Could somebody possibly explain?

Thanks a lot.