Epsilon-Delta Definition: limit, x-> -4, of (x^2 + 4x)

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nbg273

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I'm not sure how to solve this problem. I need to prove that for every Epsilon>0, there exists a Delta such that if 0<|x-a|<d, then |f(x)-L|<Epsilon.

lim (x^(2)+4x)
x-->-4
 
nbg273 said:
I'm not sure how to solve this problem. I need to prove that for every Epsilon>0, there exists a Delta such that if 0<|x-a|<d, then |f(x)-L|<Epsilon.

lim (x^(2)+4x)
x-->-4
Click to expand...
Try using the techniques displayed in your textbook in the worked examples, and by your instructor in your class notes. The first step, of course, will be the scratch-work; namely, figuring out what the limit value, L, will be. The next step will be figuring out how to relate this to "delta". And so forth.

If you're not sure how to proceed, please post one of those worked examples, and we'll see if we can't step you through the process. Thank you! ;)
 
stapel said:
Try using the techniques displayed in your textbook in the worked examples, and by your instructor in your class notes. The first step, of course, will be the scratch-work; namely, figuring out what the limit value, L, will be. The next step will be figuring out how to relate this to "delta". And so forth.

If you're not sure how to proceed, please post one of those worked examples, and we'll see if we can't step you through the process. Thank you! ;)
Click to expand...

So I kind of know what to do, but I'm stuck in the process of finding a Delta. Also, I don't know if the limit can be = 0, because if you plug -4 into x, it = 0.

If so, I started with this:
* |(x^(2)+4x)-0| < Epsilon

Then
* |(x^(2)+4x)| < Epsilon

Then I factored:
* |x(x+4)| < Epsilon

I know I'm trying to get "|x+4| < Epsilon," which I could then set to "< Delta." But I can't divide to get rid of the x on the left, because I can't have an x on the right either. This is as far as I got, that is, if everything is correct so far. But I'm worried that the limit isn't 0 and that I can't just calculate it by plugging in -4.
 
Last edited:
nbg273 said:
So I kind of know what to do, but I'm stuck in the process of finding a Delta. Also, I don't know if the limit can be = 0, because if you plug -4 into x, it = 0.

If so, I started with this:
* |(x^(2)+4x)-0| < Epsilon

Then
* |(x^(2)+4x)| < Epsilon

Then I factored:
* |x(x+4)| < Epsilon

I know I'm trying to get "|x+4| < Epsilon," which I could then set to "< Delta." But I can't divide to get rid of the x on the left, because I can't have an x on the right either. This is as far as I got, that is, if everything is correct so far. But I'm worried that the limit isn't 0 and that I can't just calculate it by plugging in -4.
Click to expand...
You know that you need to have 0 < |x - (-4)| < delta. In factored form, you have (x - (-4)) = (x + 4). So you've got a "delta" in there. (They usually set things up helpfully like this.) Now:

. . .|x (x + 4)| < (delta)|x|

...because we have to choose |x + 4| to be less than "delta". This means that:

. . .-(delta) < x + 4 < (delta)

. . .-4 - (delta) < x < -4 + (delta)

But (delta) is positive, so:

. . .-4 + (delta) < (delta)

Then:

. . .|x (x + 4)| < (delta)^2

We need:

. . .|x (x + 4)| < (epsilon)

Can you see a way to relate (epsilon) and (delta)? Once you do this, you can do your proof "front-ways", because the above (being what they don't always show you) is the backwards way -- but it's how you figure out how to do it the "right" way. ;)
 
stapel said:
You know that you need to have 0 < |x - (-4)| < delta. In factored form, you have (x - (-4)) = (x + 4). So you've got a "delta" in there. (They usually set things up helpfully like this.) Now:

. . .|x (x + 4)| < (delta)|x|

...because we have to choose |x + 4| to be less than "delta". This means that:

. . .-(delta) < x + 4 < (delta)

. . .-4 - (delta) < x < -4 + (delta)

But (delta) is positive, so:

. . .-4 + (delta) < (delta)

Then:

. . .|x (x + 4)| < (delta)^2

We need:

. . .|x (x + 4)| < (epsilon)
Click to expand...

I'm a little confused. Where did the (delta)^2 come from? And you said we need |x (x + 4)| < (epsilon), but that's what I have and I'm stuck trying to get rid of the factored "x" on the left. Maybe I didn't fully understand what you were saying.
 
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