# Need Help With Calculus Question

1. Jan 21, 2012

### CJ256

1. The problem statement, all variables and given/known data

Note: The functions above are in piece wise form, I just didn't know how to put them in piece wise function here.

2. Relevant equations

3. The attempt at a solution

http://www.mathway.com/math_image.aspx?p=SMB02LSMB03x:-1,-xSMB02ESMB032SMB02eSMB03+2SMB02lSMB03?p=87?p=38 [Broken] = 1

http://www.mathway.com/math_image.aspx?p=SMB02LSMB03x:-1,4x+5SMB02lSMB03?p=83?p=38 [Broken] = 1

1/1 = 1

Answer: limit as x approaches f(x) = 1

Last edited by a moderator: May 5, 2017
2. Jan 21, 2012

### genericusrnme

I'm not sure what your question is..

3. Jan 21, 2012

### CJ256

Edited the question! Hopefully you will be able to see it and understand it and help me out.

4. Jan 21, 2012

### genericusrnme

I understand the question in the image but I don't understand your question
I'll try and explain piecewise functions but I may be going in the wrong direction

your f(x) can be thought of as two seperate functions in the two seperate areas, when x is greater than or equal to -1 we treat the function as 4x+5
when x is less than -1 we treat the function as -x^2 + 2

In trying to find the limit, in this simple case you can just evaluate the function at x=-1. So we would use the region where x is greater than or equal to -1 and get -4+5 = 1.
I'm assuming this is all that you're expected to be doing since it seems like this is an early calculus class that isn't going to be all rigorous.

I attatched a graph of your function if it will help

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5. Jan 21, 2012

### CJ256

So the answer to my question is just limit as x approaches -1 f(x) = 1 ? And will it be acceptable for me to just show the two cases that I have in my original answer, where I have plugged in a -1 every where I have an x?

6. Jan 21, 2012

### SammyS

Staff Emeritus
While it is true that $\displaystyle \lim_{x\to-1}\,f(x)=1\,,$ what you wrote isn't quite what you need to say.

Also, what does 1/1 = 1 have to do with this question ?

You need to show that $\displaystyle \lim_{x\to-1^-}\ f(x)=\lim_{x\to-1^+}\ f(x)\,.$ If that is true then that common result is the limit you're looking for.

While it is true that $\displaystyle \lim_{x\to-1^-}\ f(x)= \lim_{x\to-1}-x^2+2\,,$ and $\displaystyle \lim_{x\to-1^+}\ f(x)= \lim_{x\to-1}4x+5\,,$ you really should state that this is what you're doing.

Last edited by a moderator: May 5, 2017
7. Jan 21, 2012

### CJ256

So can I just rewrite your last sentence and add something like:

The above statements hold true therefore limit as x approaches -1 f(x) = 1. I mean would that be an acceptable answer?

8. Jan 22, 2012

### SammyS

Staff Emeritus
As an instructor I would want you to indicate that you're taking a limit as x approaches -1 from the left, and taking a limit as x approaches -1 from the right.

That may be picky on my part, and many of my students would agree.