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

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 Charles Hottel Posts: 10 Registered: 9/5/11
Re: Limit Problem
Posted: Jan 27, 2013 12:31 PM

"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.

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