# Need Help With Calculus Question

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

## 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

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I'm not sure what your question is..

I'm not sure what your question is..

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

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

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

#### Attachments

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

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?

SammyS
Staff Emeritus
Homework Helper
Gold Member

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

## 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
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:
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.

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?

SammyS
Staff Emeritus