# Epsilon Delta Proof of an Abs. Value Function

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• Sep 13th 2013, 05:41 PM
mattjp213
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?
• Sep 13th 2013, 07:30 PM
chiro
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.
• Sep 14th 2013, 02:14 PM
mattjp213
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.
• Sep 14th 2013, 05:02 PM
chiro
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.
• Sep 14th 2013, 07:38 PM
mattjp213
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.
• Sep 14th 2013, 07:51 PM
chiro
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).
• Sep 14th 2013, 09:51 PM
mattjp213
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.
• Sep 14th 2013, 11:56 PM
mattjp213
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?