Understanding Limits & Minimum in Spivak's Calculus

[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 that's 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 \deltas
one for the x^2 and the other for x,
if 0 < |x - a| < \delta1, then |x^2 - a^2| < \epsilon/2
if 0 < |x - a| < \delta2, then |x - a| < \epsilon/2

now, for some reason i don't know,

\delta1 = min (1, \epsilon/2/2|a| + 1)
\delta2 = \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!

 

[HallsofIvy]

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 that's 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 \deltas
one for the x^2 and the other for x,
if 0 < |x - a| < \delta1, then |x^2 - a^2| < \epsilon/2
if 0 < |x - a| < \delta2, then |x - a| < \epsilon/2

now, for some reason i don't know,

\delta1 = min (1, \epsilon/2/2|a| + 1)
\delta2 = \epsilon/2

Well, to have |x-a|&lt;\epsilon/2[/tex] you must take \delta = \epsilon/2 so the that |x-a|&amp;lt; \epsilon/2 and |x-a|&amp;lt; \delta are exactly the same! \delta_1= \epsilon/2<br /> |x^2- a^2|&amp;lt; \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|&amp;lt; \epsilon/2 so we must have |x-a|&amp;lt; \epsilon/(2|x+a|) (remember we want to be able to write |x-a|&amp;lt; \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|&lt; 1 (I chose 1 just because it is easy). That means that -1&lt; x-a&lt; 1. Now add 2a to both sides. Then 2a-1&lt; x+ a&lt; 2a+ 1. For any a, 2|a|+1 is the smaller of those so we know that |x+a|&gt; 2|a|+1 and 1/|x+a|&lt; 1/(2a+1). and so |x-a|&amp;lt; (1/(2|a|+1))(\epsilon/2). Taking \delta_2= 1 guarentees that |x-a|&amp;lt; (1/(2|a|+1))(\epsilon/2).<br /> If we take \delta to be the smaller of those two, that is, take \delta1 = min (1, \epsilon/2/2|a| + 1) <b>both</b> of those are true so if |x-a|&amp;lt; \delta, <br /> |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|<br /> = (1/(2|a|+1))(\epsilon/2)|x+a|+ \epsilon/2\le \epsilon/2+ \epsilon/2= \epsilon.

 

[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?