Some calculus help?

[SleepWalkerX]

Here's the problem: lim x -> 7- f(x) = |8x - 56| / (7 - x)

Now I'm trying to figure out how to find the limit of this function as x approaches 7 from the left. For some reason I can't remember how to figure this out with this absolute value in the way. I remember something about always thinking of the absolute value of x as a step function, but I'm not sure how it would look inside the step function. I don't care if you give the answer, I just need to wrap my head around this. Any help would be appreciated.

 

[IHateMyJob2004]

doesn't this involve dirivitives?

Been 10 years ... so don't laugh if I'm wrong.

 

[Toastedlightly]

I don't think the absolute value affects anything... Both approach zero, so use L'hospital's rule (however it is spelled)

 

[Soccer55]

The absolute value function is not a step function, so you wouldn't want to look at it like that. You're probably thinking of a piecewise function, not step function. So to do what you want, you would have to figure out the equations of the lines that would make up the piecewise function and make sure you specify the domain where each part is valid. For example: f(x) = | x | can be written as a piecewise function with 2 parts. f(x) = x for x >= 0 and -x for x < 0.

-Tom

 

[SleepWalkerX]

Originally posted by: IHateMyJob2004
doesn't this involve dirivitives?

Been 10 years ... so don't laugh if I'm wrong.

I don't think so, at least we haven't started derivitives yet so I hope not. /shrugs

 

[KingGheedora]

That function is undefined for x=7, therefore the limit as x approaches 7 is undefined.

 

[SleepWalkerX]

Originally posted by: Soccer55
The absolute value function is not a step function, so you wouldn't want to look at it like that. You're probably thinking of a piecewise function, not step function. So to do what you want, you would have to figure out the equations of the lines that would make up the piecewise function and make sure you specify the domain where each part is valid. For example: f(x) = | x | can be written as a piecewise function with 2 parts. f(x) = x for x >= 0 and -x for x < 0.

-Tom

Oops, you're right I meant piecewise. I'm still kinda confused. Should I be worrying about setting up the function as a piecewise function at all?

 

[SleepWalkerX]

Originally posted by: KingGheedora
That function is undefined for x=7, therefore the limit as x approaches 7 is undefined.

But what about the limit as x approaches 7 from the left?

 

[Born2bwire]

Originally posted by: SleepWalkerX

Originally posted by: KingGheedora
That function is undefined for x=7, therefore the limit as x approaches 7 is undefined.

But what about the limit as x approaches 7 from the left?

8

 

[SleepWalkerX]

Originally posted by: Born2bwire

Originally posted by: SleepWalkerX

Originally posted by: KingGheedora
That function is undefined for x=7, therefore the limit as x approaches 7 is undefined.

But what about the limit as x approaches 7 from the left?

8

Why?

 

[BrownTown]

-8

 

[SleepWalkerX]

Originally posted by: BrownTown
-8

Explain.

 

[Born2bwire]

It's already been stated. You take it piecewise and use L'Hospital's rule. Hell, the function itself is piecewise constant. The limit is undefined because of the fact that the limit approaches different values depending upon if you take it from the left or right.

EDIT: I mean, it's really because that's what my calculator tells me.

 

[Toastedlightly]

8/7 I believe...

 

[BrownTown]

You believe wrong

 

[Toastedlightly]

Originally posted by: BrownTown
You believe wrong

I thought it followed l'hopital's rule due to both the numberator and denominator going to zero...

 

[Toastedlightly]

doh, damn absolute value!

damn me misreading 7-x as 7-7x~!

 

[Born2bwire]

Originally posted by: Born2bwire

Originally posted by: SleepWalkerX

Originally posted by: KingGheedora
That function is undefined for x=7, therefore the limit as x approaches 7 is undefined.

But what about the limit as x approaches 7 from the left?

8

Originally posted by: BrownTown
-8

Originally posted by: Toastedlightly
8/7 I believe...

/me pulls up a chair and popcorn to see how this one plays out.

 

[Toastedlightly]

Originally posted by: Born2bwire

Originally posted by: Born2bwire

Originally posted by: SleepWalkerX

Originally posted by: KingGheedora
That function is undefined for x=7, therefore the limit as x approaches 7 is undefined.

But what about the limit as x approaches 7 from the left?

8

Originally posted by: BrownTown
-8

Originally posted by: Toastedlightly
8/7 I believe...

/me pulls up a chair and popcorn to see how this one plays out.

Ends up with me admitting my mistake You aren't old enough to do things like that... go back to the kids table!

and from the left would be 8

 

[Soccer55]

Originally posted by: SleepWalkerX

Originally posted by: Soccer55
The absolute value function is not a step function, so you wouldn't want to look at it like that. You're probably thinking of a piecewise function, not step function. So to do what you want, you would have to figure out the equations of the lines that would make up the piecewise function and make sure you specify the domain where each part is valid. For example: f(x) = | x | can be written as a piecewise function with 2 parts. f(x) = x for x >= 0 and -x for x < 0.

-Tom

Oops, you're right I meant piecewise. I'm still kinda confused. Should I be worrying about setting up the function as a piecewise function at all?

Well, here's what I'd do. We know that |a*b| = |a|*|b|, so |8x-56| = |-56+8x| = |-8*(7-x)| = |-8|*|7-x| = 8*|7-x|. The reason I do that is since we're looking at this limit from the left, 7-x will always be positive. Therefore, |7-x| = 7-x. So now when you take the limit from the left, you have to look at 8*(7-x)/(7-x) = 8.

Note: |7-x| = 7-x is only true because we're looking at the limit from the left as x -> 7. If we were looking at the limit from the right, it would not be true.

-Tom

 

[ImDonly1]

I am in AP calc the answer goes like this you factor the top so its -8(-x+7) / (7-x)

they cancel out and you are left with -8.

(by the way changing the signs in an absolute value does not change it) example |7-3| = 4 and |-7+3| = 4

 

[SleepWalkerX]

Originally posted by: Toastedlightly

Originally posted by: Born2bwire

Originally posted by: Born2bwire

Originally posted by: SleepWalkerX

Originally posted by: KingGheedora
That function is undefined for x=7, therefore the limit as x approaches 7 is undefined.

But what about the limit as x approaches 7 from the left?

8

Originally posted by: BrownTown
-8

Originally posted by: Toastedlightly
8/7 I believe...

/me pulls up a chair and popcorn to see how this one plays out.

Ends up with me admitting my mistake You aren't old enough to do things like that... go back to the kids table!

and from the left would be -8

That's ok, at least you tried to help. I appreciate it.

 

[BrownTown]

derivitive of |8x - 56| = 8
derivitive if (7 - x) = -1

8/-1 = -8

That seems right, not sure if the ABS is gonna mess it though

 

[BrownTown]

Wish I had mathematica still, but my HD died and i don't know how to get my liscense back.

 

[Born2bwire]

Originally posted by: BrownTown
derivitive of |8x - 56| = 8
derivitive if (7 - x) = -1

8/-1 = -8

That seems right, not sure if the ABS is gonna mess it though

For x < 7: f(x) = 8. For x > 7: f(x) = -8.
lim(x->7-) of f(x) = 8.

You do not even need to apply L'Hospital's rule here since the function ends up being piecewise constant.