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-   -   Proof using epsilons and deltas. (http://www.physicsforums.com/showthread.php?t=6810)

gigi9 Oct6-03 06:09 PM

Proof using epsilons and deltas.
 
Can anyone please show me how to prove the problem below using epsilons and deltas. Thanks a lot...I'm so stuck...
I got through this step: /(-5x-5)/(3+x)/ < Epsilon, and don't know where to go from there. Plz help me out, thanx very much for ur valuable time.
***Prove, using epsilons and deltas, that the limit as x approaches (-1) of the function f(x) = 10/(3 + x) is 5. (Can you think of a pedagogical reason for that change?) Hint: you'll have to fiddle around and you'll probably have to finally choose your delta to be the smaller of two numbers.****

StephenPrivitera Oct6-03 06:30 PM

You have to find a d>0 such that if |x+1|<d then |f(x)-5|<e
You have been given an epsilon, you must show that there exists some d>0 corresponding to this epsilon (and for all epsilons).

So, you must find some sort of relation between e and d. Given this value of e, where must x lie? If you factor -5 from |-5x-5| you get |-5(x+1)|=5|x+1|. Try solving for |x+1|. Bear in mind that you want |x+1|<d.

HallsofIvy Oct6-03 07:08 PM

A Steven Privatera noted, you want |f(x)-5|< epsilon and you have already reduced that to |(-5x-5)/(3+x)|< epsilon which is, of course, the same as 5|x-(-1)|/|3+x|< epsilon. That is the same as
|x-(-1)|< (epsilon/5)|3+x|. The problem is that you need a constant on the right, not something that depends on x!

Okay: you know you want x close to -1. Let's for the moment assume that |x+1|< 1 (just because 1 happens to be easy to work with).

-1< x+1< 1 so -1+ 2<x+1+ 2< 1+2 or 1< |x+3|< 3. Since 1< |x+3|,
(epsilon/5)< (epsilon/5)|3+x|. (Notice the direction of the inequality.) So if we take |x-(-1)| less than the SMALLER of epsilon/5 AND 1, we have both |x+1|< epsilon/5 and epsilon/5< (epsilon/5)|3+x| so |x+1|< (epsilon/5||x+3| and you can work backwards to get 5|x+1|/|x+3|= |10/(3 + x)-5|< epsilon which is what you want.

ejdarling Oct7-03 09:31 PM

I think that you might want to show that the change in f(x) approaches zero as the change in x approaches zero. That way, the function wont do anything wild or crazy around the point at which we are calculating the limit, and will approach its calculated value in a smooth manner. To me that shows that the limit of the function as x approaches any value is the function of the value itself...

show that f(x+dx) - f(x) [the change in f(x)] is zero for dx = 0:


f(x) = 10 / (3 - x)
f(x+dx) = 10 / (3 - (x+dx))
f(x+dx) - f(x) = (10 / (3 - (x+dx))) - (10 / (3 - x))

(make common denominator):

= ((10 * (3 - x)) / (3 - (x+dx)) * (3 - x)) -
(10 * (3 - (x+dx)) / (3 - x) * (3 - (x+dx))

= ((10 * (3 - x)) - (10 * (3 - (x+dx)) /
((3 - x) * (3 - (x+dx)))
(extend all the terms):

=(30 - 10x - 30 + 10x + 10dx) /
(9 - 3x -3x - 3dx + x*x - xdx)


=10dx / (9 - 6x -3dx + x*x - xdx)

(add (3-x)dx to top & bottom):

=10dx / (9 - 6x + x*x - (3-x)dx)

=(10dx + (3-x)dx) / (9 - 6x + x*x)

=((10 + (3-x))dx) / (9 - 6x + x*x). Now, since dx is a factor in the numerator, and is eliminated from the denominator, I say that proves this expression, which is the change in f(x) corresponding to a change in x, goes to zero as dx goes to zero. I hope this helps. I'm not sure what an epsilon is; I think the delta is my dx.
-Eric

phoenixthoth Oct9-03 05:57 AM

the request was to use the epsilon delta definition of limit and not infinitesimals.

Soroban Oct9-03 11:17 AM

Hello, gigi9!

I have an answer.
(But am I oversimplifying the problem?)

We have: |10/(3 + x) - 5| < e
or: |(-5x - 5)(3 + x)| < e

and: |x - (-1)| < d
or: |x + 1| < d

The epsilon statement becomes:
|(-5)(x + 1)|/(3 + x)| < e
|-5||x + 1|/|3 + x| < e
5|x + 1|/|3 + x| < e

If x is "close" to -1, then (3 + x) is "close" to +2.
Then we have: 5|x + 1| < e/(3 + x) < e
or: |x + 1| < e/5

I believe we can use: d = e/5


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