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Help with delta arguments...

by dcl
Tags: arguments, delta
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dcl
#1
May31-04, 10:48 PM
P: 55
I'm really stuck when it comes to proving things with delta-N or delta-Epsilon arguments.

I think my biggest problem is that I don't really see how they work or 'how' they prove the 'limit' or what have you.

Would anyone be able to show me how to do some of the following questions and somewhat explain what is going on..

Use delta-N arguments to prove that:
[tex]\mathop {\lim }\limits_{n \to \infty } (n + 4)^2 =0[/tex]

Use delta-epsilon arguements to prove that as x -> 3
[tex]5x \to 15[/tex]

Guess the limit and Use delta-epsilon arguments to prove your guess correct.
[tex]
\mathop {\lim }\limits_{x \to 4} \frac{1}{{1 + x^2 }}
[/tex]

Many thanks in advance. My notes don't make it clear how I'm supposed to do this :(
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HallsofIvy
#2
Jun1-04, 06:10 AM
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PF Gold
P: 39,533
Well, first of all, you can't prove that
[tex]\mathop {\lim }\limits_{n \to \infty } (n + 4)^2 =0[/tex]
because it is not true. It should be obvious that that sequence has no limit. (If you had "-2" as the exponent instead of 2, then it would go to 0.)

The definition of "limit of a function" is "Given ε> 0, there exist a δ>0 such that if |x- x0|< δ then |f(x)- L|< ε. The "standard" proof of limits that you see in books starts from |f(x)-L|< &epsilong; and works backwards, calculating δ to show that it exists.

To show that [itex]\mathop {\lim }\limits_{n \to 3 } 5x =15[/itex] we need to get to |5x-15|< ε. Of course, |5x-15|= 5|x-3| so |5x-15|< ε is the same as |x-3|< &epsilon/5. We can take δ= &epsilon;/5.

[tex]\mathop {\lim }\limits_{x \to 4} \frac{1}{{1 + x^2 }}[/tex]
is considerably harder! First, because I know that [itex]\frac{1}{1+x^2}[/itex] is a continuous function, I would "guess" that the limit is [itex]\frac{1}{1+4^2}= \frac{1}{17}[/itex].
That means I want to arrive at [itex]|/frac{1}{1+x^2}-/frac{1}{17}|< \epsilon[/itex].

That is: [itex]|\frac{17- 1- x^2}{1+x^2}|= |\frac{x^2- 16}{1+x^2}|= |\frac{(x-4)(x+4)}{1+x^2}|= |x-4||\frac{x+4}{1+x^2}|< \epsilon[\itex].

The "x-4" term is exactly what we want. We have [itex]|x-4|< \epsilon\frac{x+4}{1+x^2}[\/tex] is close to 4, what must [itex]\frac{x+4}{1+x^2}[/itex] be close to?
arildno
#3
Jun1-04, 03:50 PM
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P: 12,016
Quote Quote by HallsofIvy

[tex]\mathop {\lim }\limits_{x \to 4} \frac{1}{{1 + x^2 }}[/tex]
is considerably harder! First, because I know that [itex]\frac{1}{1+x^2}[/itex] is a continuous function, I would "guess" that the limit is [itex]\frac{1}{1+4^2}= \frac{1}{17}[/itex].
That means I want to arrive at [itex]|\frac{1}{1+x^2}-\frac{1}{17}|< \epsilon[/itex].

That is: [itex]|\frac{17- 1- x^2}{1+x^2}|= |\frac{x^2- 16}{1+x^2}|= |\frac{(x-4)(x+4)}{1+x^2}|= |x-4||\frac{x+4}{1+x^2}|< \epsilon[/itex].

The "x-4" term is exactly what we want. We have [itex]|x-4|< \epsilon\frac{1+x^2}{x+4}[/itex]; x is close to 4, what must [itex]\frac{1+x^{2}}{x+4}[/itex] be close to?
Just removing a few beauty spots..

Gza
#4
Jun1-04, 05:01 PM
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P: 525
Help with delta arguments...

I've probably learned more math and physics from reading posts and working problems that people have difficulty with, than sitting passively in class for hours.
dcl
#5
Jun2-04, 11:17 PM
P: 55
Thanks for that, think I'm getting the hang of it....
Yeh, the first limit problem was a typo, it was meant to be:
[tex]\mathop {\lim }\limits_{n \to \infty } (n + 4)^-2 =0[/tex]
I don't suppose you could show me that, I 'think' I may have done it but I REALLY can't be sure :(
arildno
#6
Jun3-04, 01:53 AM
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You must show, for arbitrary [tex]\epsilon>0[/tex] that there exist an N so that for all n>N, [tex]\frac{1}{(n+4)^{2}}<\epsilon[/tex] :

1. [tex]\frac{1}{(n+4)^{2}}[/tex] is decreasing with n

This should be fairly easy to prove!
Hence, if you are able to find an N which satisfy [tex]\frac{1}{(N+4)^{2}}<\epsilon[/tex] (for a given [tex]\epsilon>0[/tex]), you have also shown it for n>N

2. We want to find N so that:
[tex]\frac{1}{(N+4)^{2}}<\epsilon[/tex]

This inequality is equivelent to:
[tex]\frac{1}{\epsilon}<(N+4)^{2}[/tex]

Hence, we find the requirement on N:
[tex]N>\sqrt{\frac{1}{\epsilon}}-4[/tex]
dcl
#7
Jun3-04, 02:24 AM
P: 55
Nice. I'm suprised it turned out to be that simple...


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