# Limits and minimum

1. Dec 31, 2007

### uak01

hi all,
like many here i have also picked up the spivak calculus text and run into some problems.
i am reading the fifth chapter which introduces 'limits' and i can't get my head around the idea of using minimum (the concept was introduced in the problems of the first chapter and it reappears in chapter 5 without any formal discussion about it). i haven't figured out how to approach it yet. i guess thats what i mean to ask here. i don't think i am making any sense. probably an example will help. this is from the book, towards the end of the chapter, consider:

f(x) = x^2 + x, and x approaches a,
according to the definition in the book, we have to find a $$\delta$$ > 0 such that |x^2 + x - (a^2 + a)| < $$\epsilon$$
it then breaks down the function into x^2 and x, so that we need 2 $$\delta$$s
one for the x^2 and the other for x,
if 0 < |x - a| < $$\delta$$1, then |x^2 - a^2| < $$\epsilon$$/2
if 0 < |x - a| < $$\delta$$2, then |x - a| < $$\epsilon$$/2

now, for some reason i don't know,

$$\delta$$1 = min (1, $$\epsilon$$/2/2|a| + 1)
$$\delta$$2 = $$\epsilon$$/2

what is going on here?

this is a very confusing first post and i am very sorry about it. (imagine the confusion in my head perhaps i can try to clarify my question after a few responses. thanks in advance!

2. Jan 1, 2008

### HallsofIvy

Staff Emeritus
Well, to have $|x-a|<\epsilon/2[/tex] you must take [itex]\delta = \epsilon/2$ so the that $|x-a|< \epsilon/2$ and $|x-a|< \delta$ are exactly the same! $\delta_1= \epsilon/2$
$|x^2- a^2|< \epsilon/2$ is a little harder. $|x^2- a^2|= |(x-a)(x+a)|= |x-a||x+a|$. In order to have $|x^2- a^2|= |x-a||x+a|< \epsilon/2$ so we must have $|x-a|< \epsilon/(2|x+a|)$ (remember we want to be able to write $|x-a|< \delta$ so we solve for |x-a|). Of course, we need that number on the right to be a constant- not depend of a. We are talking about x close to a so suppose we take |x-a|< 1 (I chose 1 just because it is easy). That means that -1< x-a< 1. Now add 2a to both sides. Then 2a-1< x+ a< 2a+ 1. For any a, 2|a|+1 is the smaller of those so we know that |x+a|> 2|a|+1 and 1/|x+a|< 1/(2a+1). and so $|x-a|< (1/(2|a|+1))(\epsilon/2)$. Taking $\delta_2= 1$ guarentees that $|x-a|< (1/(2|a|+1))(\epsilon/2)$.
If we take $\delta$ to be the smaller of those two, that is, take $$\delta$$1 = min (1, $$\epsilon$$/2/2|a| + 1) both of those are true so if $|x-a|< \delta$,
$|x^2+x-(a^2+a)|= |(x^2-a^2)+ (x-a)|\le |x^2- a^2|+ |x-a|= |x-a||x+a|+ |x-a|$
$= (1/(2|a|+1))(\epsilon/2)|x+a|+ \epsilon/2\le \epsilon/2+ \epsilon/2= \epsilon$.

3. Jan 1, 2008

### uak01

thanks for the response HallsofIvy, I think I follow it mostly. one more small query: why must we take the minimum to express delta? my uninformed guess is that it is to satisfy the absolute values of |x - a| form or, because of the power of x (x^2, x^3, etc).
what do you think?