# Find values of a for which limit exists (piecewise function)

• Aug 31st 2010, 06:19 PM
yzobel
Find values of a for which limit exists (piecewise function)
f(x)=2-x, x<-1
x, -1<=x<1
(x-3)^2, x>=1

Find all the values of a for which lim(x approaches a) f(x) exists. Express answer in interval notation.

Please correct me if I'm wrong. I think that a can exist for the first function becase the limits approaching from the left and right are the same. Same goes for the last function. But what about the middle function?

Thank you, I appreciate any help!
• Aug 31st 2010, 07:31 PM
mr fantastic
Quote:

Originally Posted by yzobel
f(x)=2-x, x<-1
x, -1<=x<1
(x-3)^2, x>=1

Find all the values of a for which lim(x approaches a) f(x) exists. Express answer in interval notation.

Please correct me if I'm wrong. I think that a can exist for the first function becase the limits approaching from the left and right are the same. Same goes for the last function. But what about the middle function?

Thank you, I appreciate any help!

Try drawing a graph of y = f(x) before answering the question.
• Aug 31st 2010, 07:48 PM
Prove It

$f(x) = \begin{cases}2-x,x < -1\\x,-1\leq x <1\\ (x-3)^2,x \geq 1\end{cases}$

All three pieces are polynomials, that are continuous everywhere except at the endpoints. So the only places where the limits might not exist is at the endpoints.

So we have to check the left and right hand limits at the endpoints. If they are equal, then the limit exists.

$\lim_{x \to -1^{-}}f(x) = 2 - (-1) = 3$.

$\lim_{x \to -1^{+}}f(x) = -1$.

Since the left and right hand limits are not the same, the function does not have a limit at $x = -1$.

$\lim_{x \to 1^{-}}f(x) = 1$.

$\lim_{x \to 1^{+}}f(x) = (1-3)^2 = 4$.

Since the left and right hand limits are not the same, the function does not have a limit at $x = 1$.

Thus, the values of $a$ for which the function has a limit are $x \in (-\infty, -1)\cup (-1, 1)\cup (1, \infty)$.
• Aug 31st 2010, 09:19 PM
yzobel