Register to reply

Help with proving limits using Epsilon-Delta definition

by shirewolfe
Tags: definition, epsilondelta, limits, proving
Share this thread:
shirewolfe
#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.
Phys.Org News Partner Science news on Phys.org
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds
Muzza
#2
Oct3-04, 08:02 AM
P: 696
Use the triangle inequality perhaps, |x + 1| = |x - 1 + 2| <= |x - 1| + 2 < d + 2.
arildno
#3
Oct3-04, 10:23 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
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
#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
#5
Oct4-04, 04:47 PM
Sci Advisor
HW Helper
PF Gold
P: 12,016
I was perhaps a bit too snappish on that point, don't bite me back, though..


Register to reply

Related Discussions
Epsilon delta limits. Calculus 39
Delta epsilon limits Calculus & Beyond Homework 6
Epsilon Delta Limits Calculus & Beyond Homework 10
Using the Delta-epsilon definition but for two variables? Calculus & Beyond Homework 3
Epsilon and delta definition of limit Calculus 4