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Topic: Limit Problem
Replies: 3   Last Post: Jan 27, 2013 12:31 PM

 Messages: [ Previous | Next ]
 William Elliot Posts: 2,637 Registered: 1/8/12
Re: Limit Problem
Posted: Jan 27, 2013 4:22 AM

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.
>
>
>
>

Date Subject Author
1/26/13 Charles Hottel
1/26/13 gnasher729
1/27/13 William Elliot
1/27/13 Charles Hottel