help with proving limits using Epsilon-Delta definition


by shirewolfe
Tags: definition, epsilondelta, limits, proving
shirewolfe
shirewolfe is offline
#1
Oct3-04, 07:25 AM
P: 3
I am having trouble proving the limits of quadratic functions such as the following. (I used "E" to represent epsilon and "d" for delta)

lim [(x^2)+1]
x->1

I found the limit, L, to equal 2 and have proceeded through the following steps:

|f(x) - L| < E
| [(x^2)+1] - 1| < E
|[(x^2)-1]| < E
|x+1||x-1| < E

While I also know that 0 < |x-1| < d.

My question is how do you find the numerical relationship between |x+1||x-1| and |x-1| so that I may find d in terms of E?

(I was thinking of finding the bounded open interval in which |x+1||x-1| = |[(x^2)-1]| and substituting the greastest figure of the interval, which would be greater than |x+1|, in place of |x+1| so that I would have z|x-1|< E where z is an identified numerical value. However, in problems like the one above, |x+1||x-1| = |[(x^2)-1]| within a seemingly unbounded interval.)

please help. your time and assistance is very much appreciated.
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Muzza
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#2
Oct3-04, 08:02 AM
P: 696
Use the triangle inequality perhaps, |x + 1| = |x - 1 + 2| <= |x - 1| + 2 < d + 2.
arildno
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#3
Oct3-04, 10:23 AM
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Quote Quote by shirewolfe
I am having trouble proving the limits of quadratic functions such as the following. (I used "E" to represent epsilon and "d" for delta)

lim [(x^2)+1]
x->1

I found the limit, L, to equal 2 and have proceeded through the following steps:
First of all, you have NOT found the limit, and certainly not shown that it is 2!
What you have done, is to make a GUESS at the limit value!!
(As it happens, you've made a correct guess, but that is irrelevant; it's still a guess).

What you are to do now, is:
1. Does my guess (2) satisfy the properties that a limit must have?
2. Let [tex]|x-1|<\delta[/tex]
3. Then, [tex]|x^{2}+1-2|=|x^{2}-1|=|x-1||x+1|[/tex]
4. Now, by assumption, [tex]|x-1|<\delta[/tex]
Let us make a further assumption, that [tex]\delta<1[/tex]
Then, [tex]|x+1|<2 (\delta<1)[/tex]
And:
[tex]|x^{2}+1-2|<2\delta, (\delta<1)[/tex]
5. Let [tex]\epsilon>0[/tex]
If we are to have [tex]|x^{2}+1-2|<\epsilon[/tex] for all x satisfying [tex]|x-1|<\delta[/tex] it is sufficient if the following inequalities are simultaneously satisfied:
[tex]2\delta<\epsilon[/tex]
[tex]\delta<1[/tex]

Hence, setting [tex]\delta=min(\frac{\epsilon}{2},1)[/tex] suffices.

That is, we were able to show that our guess (2) satisfy the properties a limit must have.

shirewolfe
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#4
Oct4-04, 04:43 PM
P: 3

help with proving limits using Epsilon-Delta definition


By 'finding' 2 through the substitution of 1 in f(x)=[(x^2)+1] I meant that it was a possible limit whose validity must be varified through the E-d definition. Thank you for correcting me however, to avoid my own future confusion.

Thanks for the helpful guidance and clarification.
arildno
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#5
Oct4-04, 04:47 PM
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I was perhaps a bit too snappish on that point, don't bite me back, though..


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