Finding Delta Algebraically Thomas' Calculus

  • #1
3,003
2

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!
 

Answers and Replies

  • #2
867
0
I'd say that since the radicals are never negative, x can't be negative so you ignore the negative part and you can drop the absolute values. Don't know about the last thing.
 
  • #3
Avodyne
Science Advisor
1,396
87
In the last line, he is specifically looking to construct an open interval of x that contains x=2. Since 2 is positive, he drops the absolute value signs. By inspection, the stated internal contains x=2.
 

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