# Thread: Limts True or False Help

1. ## Limts True or False Help

Hey all here is an easy calc. question but i can't seem to understand how to answer it

Question:

True or False? If the statement is true, explain why; if it is false, then explain how to determine the value of $\displaystyle c$ and then, evaluate the limit.

There is no number $\displaystyle c$ such that $\displaystyle \lim_{x \rightarrow\ -1} \frac{x^2 + c·x - c + 2}{x^2 - 1}$ exists.

The Part that I need help with:
I know that answer is false but how do i evaluate $\displaystyle c$?

2. Originally Posted by jospams
Hey all here is an easy calc. question but i can't seem to understand how to answer it

Question:

True or False? If the statement is true, explain why; if it is false, then explain how to determine the value of $\displaystyle c$ and then, evaluate the limit.

There is no number $\displaystyle c$ such that $\displaystyle \lim_{x \rightarrow\ -1} \frac{x^2 + c·x - c + 2}{x^2 - 1}$

The Part that I need help with:
I know that answer is false but how do i evaluate $\displaystyle c$?

There is no statement here to be false.

3. Originally Posted by Drexel28
There is no statement here to be false.
are you sure(someone in the class said it was false and i have to factor something to get the answer)? why is it true?

4. The limit can only exist if there is a factor of (x+1) in the numerator to cancel with the (x+1) term in the denominator. If such a factor exists in the numerator, then the discontinuity at x=-1 will be removable (meaning the limit does exist) instead of infinite (meaning the limit does not exist).

Therefore we must find what value of c causes $\displaystyle x^2+cx-c+2$ to be divisible by $\displaystyle x+1$. You can do this by using long division to find the remainder, and setting the remainder equal to zero.

5. If you want to avoid long division, another method is to just set up the equality:

$\displaystyle x^2+cx-c+2 = (x+1)(x+k)$ where k is some constant...

Therefore:

$\displaystyle x^2+cx-c+2 = x^2+(k+1)x+k$

So,

$\displaystyle c=k+1$ and $\displaystyle -c+2=k$

Solving this gives us $\displaystyle c=3/2$ (which is the same answer I got with long division method).

6. Originally Posted by Drexel28
There is no statement here to be false.
Yes, there was. The statement was "There is no number c such that ...".

(Unless it was edited to change the statement before I got here.)

7. Originally Posted by HallsofIvy
Yes, there was. The statement was "There is no number c such that ...".

(Unless it was edited to change the statement before I got here.)
Exists wasn't there before. It just said

"There is no $\displaystyle c$ such that $\displaystyle \lim\cdots$"