# Finding Delta Algebraically Thomas' Calculus

1. Sep 7, 2009

1. The problem statement, all variables and given/known data

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

Problem:

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

Solution

Step 1 Solve the inequality $$|f(x)-4|<\epsilon$$ to find an open interval containing xo = 2 on which the inequality holds for all $x\ne x_0$

For $x\ne x_0 = 2$, we have f(x) = x2 and the inequality to solve is
$$|x^2-4|<\epsilon$$

Thus, (Author's comments in Blue) :

$$|x^2-4|<\epsilon$$

$$-\epsilon < x^2-4<\epsilon$$

$$4-\epsilon < x^2 < 4 + \epsilon$$

$$\sqrt{4-\epsilon} < |x| < \sqrt{4 + \epsilon}$$ Assumes $\epsilon < 4$

$$\sqrt{4-\epsilon} < x < \sqrt{4 + \epsilon}$$ 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!

2. Sep 7, 2009

### Bohrok

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. Sep 7, 2009

### Avodyne

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.