Find delta from a particular epsilon

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la_med12
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I'm learning how to find delta from a particular epsilon. I'm not understanding a step in the solution for the problem listed below:

Here's the problem:
lim(x→3)x^2=9

Solution:
|x^2-9|<.05
-.05<x^2-9<.05
2.9916..<x<3.0083
2.9916..-3<x-3<3.0083..-3
-.0083..<x-3<.0083..
delta=.0083

I don't understand how:
8.95<x^2<9.05
calculates to:
2.9916..<x<3.0083..
Could someone please explain this to me.. in simple terms. Cheers!
 
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Take the square root of each member.
 
Ah.. I guess I was looking at it for so long that I overlooked the obvious. Thanks for the help!
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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