Date: Jan 27, 2013 12:31 PM
Author: Charles Hottel
Subject: Re: Limit Problem

"William Elliot" <marsh@panix.com> wrote in message

news:Pine.NEB.4.64.1301261746060.3500@panix3.panix.com...

> On Sat, 26 Jan 2013, Charles Hottel wrote:

>

>> I am having a problem following an example in my book.

>>

>> Prove lim(x->c) 1/x = 1/c, c not equal zero

>>

> Assume c /= 0, |x - c| < s, s < c

>

> -s < x - c < s

> c - s < x < c + s

>

> 1/(c + s) < 1/x < 1/(c - s)

> 1/(c + s) - 1/c < 1/x - 1/c < 1/(c - s) - 1/c = (c - c + s)/c(c - s)

>

> -s/c(c - s) < 1/x - 1/c < s/c(c - s)

> |1/x - 1/c| < s/c(c - s)

>

> (c^2 - cs)r = s

> Let s = rc^2 / (1 + cr)

>

> s/c(c - s) = [rc^2 / (1 + cr)] / c(c - rc^2 / (1 + cr))

> rc^2 / c(c + rc^2 - rc^2) = r

>

> Given r > 0, take s as above to show

> |x - c| < s implies |1/x - 1/c| < r.

>

> Be sure to make s small enough so that s < c.

>

>> So 0 < | x-c| < delta, implies |1/x - 1/c| < epsilon

>>

>> |1/x - 1/c| = | (c-x) / {xc}| = 1/|x| * 1/|c| * (x-c) < epsilon

>>

>> Factor 1/|x| is troublesome if x is near zero, so we bound it to keep it

>> away from zero.

>>

>> So |c| = |c - x + x| <= |c-x| + |x| and this imples |x| >= |c| - |x-c|

>>

>> I think I understand everything up to this point, but not the next

>> steps,

>> which are

>>

>> If we choose delta <= |c|/2 we succeed in making |x| >= |c| / 2.

>> Finally if we require delta <= [(epsilon) * (c**2)} / 2 then

>>

>> [1/|x| * 1/|c| * |x-c|] < [1 / (|c|/2)] * [1/|c|] * [((epsilon) *

>> (c**2)) / 2] = epsilon

>>

>> How did they know to choose delta <= |c|/2?

>>

>> How does that lead to |x| > |c|/2 implies 1/|x| < 1/(|c|/2) ?

>>

>> I did not sleep well last night and I feel I must be missing something

>> that would be obvious if my head was clearer. Thanks for any help.

>>

>>

>>

>>

Thanks, I had sort of figured it out on my own but you post makes it

clearer.