# Thread: Epsilon Delta Proof of an Abs. Value Function

1. ## Epsilon Delta Proof of an Abs. Value Function

I am having trouble with this epsilon delta problem.

Prove that the limit as x approaches -5 of f(x) = |x-5| is 10.

I am confused because |f(x)-L| < e gives us ||x-5|-10| < e and I'm not sure how to handle the inside absolute value.

My thought is that since there are two cases here - x-5 being positive or negative, you can solve for each w/o the
absolute value brackets.

If x-5 > 0, ||x-5|-10| = |x-5-10| = |x-15| < e

If x-5 < 0, ||x-5|-10| = |x-5+10| = |x+5| < e (This one looks like the d inequality 0 < |x+5| < d)

All I can think of is to say that since they are both equivalent after the absolute value is taken that you can use either simplification to choose an e based d to plug in for the proof. So you would choose the one that looks like the d inequality to say that if you choose a d = e and if 0 < |x+5| < d, then |f(x)-10| = |x+5| < d which = e. Proof done.

I don't feel good about just choosing the version of f(x) - 10 that I like and throwing the other away. Is this correct or am I off track here?

2. ## Re: Epsilon Delta Proof of an Abs. Value Function

Hey mattjp213.

Hint: Break up the cases (you will have four cases) for ||x-5|-10|. The first case is 5-x when x < 5 and x-5 when x >= 5. Now you have |5-x-10| and |x-5-10| and you need to figure out when this expresion is < 0 and >= 0 which will be related to when |-x-5| and |x-15| are whatever they are.

3. ## Re: Epsilon Delta Proof of an Abs. Value Function

But if I simplify |x+5| and |x-15| into the four cases that equal their absolute value to get rid of the | | operator then I think I will have gone too far for this problem.

Solving the four cases gives:
|x+5| when x<-5, = -x-5 and when x>=-5, = x+5.
|x-15| when x<15, = -x+15 and when x>=15, = x-15.

So I can solve for the four cases, but I thought I had to leave the last absolute value intact so it can compare to 0<|x+5|<d.
I thought I only had to get rid of the nested absolute value in ||x-5|-10|<e, but you're asking me to go further. I am not sure if you misunderstood what I was exactly trying to do with this problem or if I just don't see how full simplification is necessary to reach the answer. I am still trying to find out how to use |f(x)-10| = |x+5| or |x-15| to prove the limit of f(x) as x approaches -5.
To restate, my confusion is in what to do with the two cases for this because of the absolute value. The first one looks like d which is what is what I want so I can find my d in relation to epsilon and plug that into my proof. The other does not look like d and I am wondering if I need to solve for d using both cases or if I can pick the one that works because they are both equal to each other when absolute value is taken. That's a lot of explanation. Hope it's not too messy.

4. ## Re: Epsilon Delta Proof of an Abs. Value Function

You just need to consider the cases in the neighborhood of x = 5. You have four cases but you only need to look at the ones relevant to the neighborhood of x = 5.

5. ## Re: Epsilon Delta Proof of an Abs. Value Function

Of the four cases, I can see that the ones applying to x = 5 would be |x+5| when x>=-5 which would = x+5 and |x-15| when x<15 which would = -x+15. I still cannot see where you're going with this. I don't see how this is bringing me closer to the limit proof or what the significance is of identifying the cases relevant to the neighborhood including x = 5. Shouldn't I be worried about the cases where x is near -5 since that is what x is approaching in my limit?

Are you trying to get me to figure out that these two cases are = to each other at x = 5? I can also see that the y value of these two at x=5 is equal to the limit of my original function (y=10). I graphed both cases and I can see that my original |f(x)-10| is equal to the graph of x+5 on [-5,5] and -x+15 on [5,15). Is this justification to substitute |f(x)-10| for either of these? It still doesn't make sense to me because these two cases don't have anything to do with the interval in which x approaches c.

6. ## Re: Epsilon Delta Proof of an Abs. Value Function

The idea is to show that the epsilon-delta proof works for the relevant cases. So if you approach it from just before and just after x = -5 then the limit should be 10.

What functions (in terms of analytic expressions not involving absolute signs) do you have for f(x) just before x = -5 and f(x) just after x = -5?

Once you have these then you use this f(x) to show both a left hand limit and a right hand limit and then you're done.

In summary you have f(x) for x < -5 (but in the neighborhood of x = -5) and f(x) for x > -5 (but again in the neighborhood of x = -5 but this time to the right).

7. ## Re: Epsilon Delta Proof of an Abs. Value Function

That makes a lot more sense. Sorry you had to spell it out for me. I'll try to put the whole thing together and post it to see if my solution makes sense.

8. ## Re: Epsilon Delta Proof of an Abs. Value Function

Here's what I have:

For: lim as x approaches -5 of |x-5|, Find L, then use the e-d definition to prove that the limit is L.

This function is continuous for all x near c so we can just plug c into the function to get L:
L = |-5-5| = |-10| = 10

Then to prove the limit is L we must show that for any given e there is a d such that if 0<|x-c|<d, then |f(x)-L|<e

If 0<|x-(-5)|<d or 0<|x+5|<d, then ||x-5|-10|<e

First we must break up |x-5| into separate cases based on the value of x.

|x-5| = 5-x when x<5 and x-5 when x>5. Then we put each of these into our |f(x)-L|<e inequality.

This gives us: |5-x-10| which equals |-x-5|, and |x-5-10| which equals |x-15|.

Each of these cases has two cases of its own:
|-x-5| = -x-5 when x<-5 and x+5 when x>-5, |x-15| = x-15 when x>15 and -x+15 when x<15

Since we are only interested in the cases when x is approaching c which is -5 here, we are only required to show that our d and
e inequalities hold true for cases of |f(x)-10| close to x=-5. This would be |f(x)-10| = -x-5 when x<-5 and x+5 when x>-5.

1st case: x<-5 or as x approaches c from the left. 2nd case: x>-5 or as x approaches c from the right.
|f(x)-10| = |-x-5| < e |f(x)-10| = |x+5| < e
-1|x+5| < e or |x+5| < e

So for both cases, if we choose d = e, then if our x is within d units of c, then f(x) will be within e units of L. Proved.

Correct?

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