# Thread: Need help with a simple continuity problem.

1. ## Need help with a simple continuity problem.

f(x) = { cos(x-2)+1 x is less than or equal to 2
{ sin(x-2)+2 2<x

Is it continuous at x=2 at the right, left, both, or neither?

I've tried to work it out and graph but i'm pretty confused right now. I think that it's not continuous because both pieces are not equal, but I dont know if it's right, left, or none.

2. ## Re: Need help with a simple continuity problem.

Originally Posted by poipoipoi10
f(x) = { cos(x-2)+1 x is less than or equal to 2
{ sin(x-2)+2 2<x

Is it continuous at x=2 at the right, left, both, or neither?

I've tried to work it out and graph but i'm pretty confused right now. I think that it's not continuous because both pieces are not equal, but I dont know if it's right, left, or none.
$\lim_{x \to 2^-} \cos(x-2)+1 = 2$

$\lim_{x \to 2^+} \sin(x-2)+2 = 2$

$f(2) = 2$

what's that tell you?

3. ## Re: Need help with a simple continuity problem.

Originally Posted by poipoipoi10
f(x) = { cos(x-2)+1 x is less than or equal to 2
{ sin(x-2)+2 2<x

Is it continuous at x=2 at the right, left, both, or neither?
$f(x) = \left\{ {\begin{array}{rl} {\cos (x - 2) + 1,} & {x \leqslant 2} \\ {\sin (x - 2) + 2,} & {x > 2} \\ \end{array} } \right.$

Here is some analysis notation: $f(a + ) = \lim _{x \to a^ + } f(x)$.
There is of course a similar statement for $f(a-)$.

Now you need to see if $f(2-)=f(2+)$.