TITLE: (Stuck on the last step) Find the largest δ such that |f(x) − L| < 0.01 whenever 0 < |x − 5| < δ
QUESTION [0 upvotes]: The problem I'm working with is this:

Consider the following limit:
$\lim\limits_{x \to 5} x^2+4$
Find the largest δ such that |f(x) − L| < 0.01 whenever 0 < |x − 5| < δ. 
(Assume 4 < x < 6 and δ > 0. Round your answer to four decimal places.)

The answer given by the book is 0.001, but I don't understand how to get there.
These are my steps so far:
1) Plug 5 into f(x) to find L.
Result: 29.
2) Plug everything into |f(x)-L|<0.01.
Result: |($x^2$ + 4) - 29| < 0.01
3) Simplify: |$x^2$ - 25| < 0.01
This is where I get stuck.
I know I can further simplify to |x-5||x+5|<0.01, but I still don't see any way of getting 0.001 from here. I must have gone wrong somewhere earlier, right? But I don't know where. 
I'm an idiot. This is my second time taking calculus and my book and the online tutorials have so many examples where everything makes perfect sense to me until they do some algebraic magic, and I have no idea why they did something or how I should know when it can be applied to future problems. Can someone please help me, and walk me through why each step is necessary.

REPLY [0 votes]: $$|x^2-25|<0.01\tag{1}$$
$$-\frac{1}{100}<x^2-25<\frac{1}{100}$$
You should be able to solve these two quadratic inequalities and get the following result:
$$x^2-25<\frac{1}{100} \implies -\frac{\sqrt{2501}}{10}<x<\frac{\sqrt{2501}}{10}$$
$$x^2-25>-\frac{1}{100} \implies x<-\frac{1}{10} \left(7 \sqrt{51}\right)\lor x>\frac{7 \sqrt{51}}{10}$$
...or when you introduce the fact that $x$ must be positive:
$$0<x<\frac{\sqrt{2501}}{10}$$
$$x>\frac{7 \sqrt{51}}{10}$$
The final solution for $x$ must satisfy both inequalities:
$$\frac{7 \sqrt{51}}{10}<x<\frac{\sqrt{2501}}{10}$$
$$\frac{7 \sqrt{51}}{10}-5<x-5<\frac{\sqrt{2501}}{10}-5$$
$$|x-5|<\delta=\min(5-\frac{7 \sqrt{51}}{10}, \frac{\sqrt{2501}}{10}-5)$$
Now grab the calculator and calculate the value of $\delta$.
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