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## Homework Statement

This is Example 5 in Chapter 2.3 of the above mentioned text:

*Problem*:

Prove that the [tex]\lim_{x\rightarrow2}f(x)=4[/tex] if [tex]f(x)= x^2 \text{ for }x\ne2\text{ and }f(x)=1\text{ for }x=2[/tex]

*Solution*

__Step 1__Solve the inequality [tex]|f(x)-4|<\epsilon[/tex] to find an open interval containing x

_{o}= 2 on which the inequality holds for all [itex]x\ne x_0[/itex]

For [itex]x\ne x_0 = 2[/itex], we have f(x) = x

^{2}and the inequality to solve is

[tex]|x^2-4|<\epsilon[/tex]

Thus, (Author's comments in Blue) :

[tex]|x^2-4|<\epsilon[/tex]

[tex]-\epsilon < x^2-4<\epsilon[/tex]

[tex]4-\epsilon < x^2 < 4 + \epsilon[/tex]

[tex]\sqrt{4-\epsilon} < |x| < \sqrt{4 + \epsilon}[/tex] Assumes [itex]\epsilon < 4[/itex]

[tex]\sqrt{4-\epsilon} < x < \sqrt{4 + \epsilon}[/tex] An open interval about x

_{o}=2 that solves the inequality.

I will stop here, though I have more questions. My initial 2 questions are these:

1.) Maybe it's obvious, but in the last step, how did he drop the abs value sign?

and

2.) I fail to understand the last blue comment. How does the preceding procedure take into account that x

_{o}= 2 ?

Thanks!