# Demonstration analysis and good books?

1. May 2, 2011

### fgyamauti

When I try to demonstrate that lim (x^2)=4
x--->2

I found a different delta (delta=min{2-sqrt(epsilon-4),sqrt(epsilon+4)-2}, towards the one that is written in Demidovich´s book (delta=epsilon/5). Could someone help me?
Could someone tell me, too, a good algebra, calculus and real analysis book? Is it ok if i try analysis next semester absent a more advanced calculus (without calculus 2)? I need some books that explain polar coordinates too (good ones for beginner).
Thanks and sorry for my mathematical notation.

Last edited: May 2, 2011
2. May 3, 2011

### HallsofIvy

Staff Emeritus
You want $|x^2- 4|= |(x- 2)(x+ 2)|= |x+ 2||x- 2|< \epsilon$.
So we can write $|x- 2|< \epsilon/|x+ 2|$
But we need a value, $\delta$ that does NOT depend on x so that is $|x- 2|< \delta$ then $|x- 2|< \epsilon/|x+ 2|$. That means we want $\delta< \epsilon/|x+2|$.

If |x- 2|< 1 (The "1" is chosen just because it is simple. Any positive number would do.) then -1< x- 2< 1 so 1< x< 3 and then 3< x+2< 5 so that 3< |x+ 2|< 5. Then
$$\frac{1}{5}< \frac{1}{|x+2|}< 1/3$$
and so
$$\frac{\epsilon}{5}< \frac{\epsilon}{|x+2|}$$

That is, if
$$|x- 2|< \frac{\epsilon}{5}$$
and |x- 2|< 1 as assumed above to get this, we will have
$$|x-2|< \frac{\epsilon}{|x+2|}$$
so
$$|x-2||x+2|= |x^2- 4|< \epsilon$$
as need.

In other words, take $\delta$ to be the smaller of $\epsilon/5$ or 1.

Analysis is basically the theory behind the Calculus. I would NOT take an analysis course without having completed the Calculus sequence.