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## #1 2021-04-05 01:14:02

mathland
Banned
Registered: 2021-03-25
Posts: 438

### Epsilon-delta Limit Proof

Use the epsilon-delta method to show that the limit is 3/2 for the given function.

lim    (1 + 2x)/(3 - x) = 3/2
x-->1

I want to find a delta so that | x - 1| < delta implies |f(x) - L| < epsilon.

| (1 + 2x)/(3 - x) - (3/2) | < epsilon

-epsilon < (1 + 2x)/(3 - x) - 3/2 < epsilon

I now add 3/2 to all terms.

(3/2) - epsilon < (1 + 2x)/(3 - x) < (3/2) + epsilon

Stuck here....

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## #2 2021-04-05 19:42:30

Bob Registered: 2010-06-20
Posts: 9,128

### Re: Epsilon-delta Limit Proof

hi mathland,

This is not a topic I have worked on before, so I've waited a while in the hope someone else might give an answer.

As that hasn't happened yet I've started by researching the method.  I found this pdf which I found very clear.

https://math.berkeley.edu/~willij/1a/epsilonics.pdf

Then I did the following.  You'll have to decide if it is a proper solution:

As x is close to 1 we may assume that 3-x > 0 therefore

So for any epsilon I have found an epsilon dependant delta .  I think that is what the method requires.

Let me know what you think.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob Offline

## #3 2021-04-06 10:07:26

mathland
Banned
Registered: 2021-03-25
Posts: 438

### Re: Epsilon-delta Limit Proof

Bob wrote:

hi mathland,

This is not a topic I have worked on before, so I've waited a while in the hope someone else might give an answer.

As that hasn't happened yet I've started by researching the method.  I found this pdf which I found very clear.

https://math.berkeley.edu/~willij/1a/epsilonics.pdf

Then I did the following.  You'll have to decide if it is a proper solution:

As x is close to 1 we may assume that 3-x > 0 therefore

So for any epsilon I have found an epsilon dependant delta .  I think that is what the method requires.

Let me know what you think.

Bob

Hello Bob. Thanks for the link and effort. The proof is more intense, more rigorous than your reply. I posted the problem at another forum. Someone was kind enough to work out the entire proof.

Scary stuff. Thank God I am NOT taking Calculus 1 in a classroom setting. I like the epsilon-delta proof but it just hasn't sinked deep enough for me to feel comfortable solving problems.

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