- #1
Saladsamurai
- 3,020
- 7
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 xo = 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) = x2 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 xo=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 xo = 2 ?
Thanks!