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

 Messages: [ Previous | Next ]
 gnasher729 Posts: 419 Registered: 10/7/06
Re: Limit Problem
Posted: Jan 26, 2013 7:35 PM

On Jan 26, 10:56 pm, "Charles Hottel" <chot...@earthlink.net> wrote:
> I am having a problem following an example in my book.
> I understand the concept of limit but sometimes I get confused
> manipulating expressions with absolute values in them.  Here is the problem:
>
> Prove  lim(x->c) 1/x  = 1/c, c not equal zero
>
> 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.

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

Once delta is chosen, you will need to examine all x with |x - c| <
delta and show that 1/x is close to 1/c.

You want to choose delta so that x stays far away from zero.

Let's say you picked delta = 0.01. If for example c = 1.43, then delta
= 0.01 means |x - 1.43| < 0.01, or 1.42 < x < 1.44. That's far away
from x = 0. But the same delta with c = 0.012 wouldn't be very good,
because delta = 0.01 now means that |x - 0.012| < 0.01, or 0.002 < x <
0.022. If c = 0.005 then this delta is awful, because |x - 0.005| <
0.01 means -0.005 < x < 0.015; even x = 0 would be included.

What happens if we pick delta = |c| / 1000? It means that x has to be
very, very close to c. Actually, x must be between 0.999c and 1.001c.
We don't know if c is positive or negative, so we can't just say
0.999c < x < 1.001 c because that would be wrong if c is negative. But
we can say that 0.999 |c| < |x| < 1.001 |c|. Similar, if we pick delta
= |c| / 2 then x must be between 0.5 c and 1.5 c, or 0.5 |c| < |x| <
1.5 |c|.

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