# Is the following function continuous at x=3 ?

## $f \left(x\right) = \left\{\begin{matrix}2 & \text{ if " x=3 \\ x-1 & " if " x > 3 \\ (x+3)/3 & " if } x < 3\end{matrix}\right.$

Mar 4, 2017

Yes

#### Explanation:

Given:

$f \left(x\right) = \left\{\begin{matrix}2 & \text{ if " x=3 \\ x-1 & " if " x > 3 \\ (x+3)/3 & " if } x < 3\end{matrix}\right.$

We find:

${\lim}_{x \to 3 -} f \left(x\right) = {\lim}_{x \to 3 -} \frac{x + 3}{3} = \frac{\textcolor{b l u e}{3} + 3}{3} = 2$

${\lim}_{x \to 3 +} f \left(x\right) = {\lim}_{x \to 3 +} \left(x - 1\right) = \textcolor{b l u e}{3} - 1 = 2$

$f \left(3\right) = 2$

So the left and right limits agree and are equal to $f \left(3\right)$.

So this $f \left(x\right)$ is continuous at $x = 3$

graph{((x-3)/abs(x-3)+1)/2(x-1)+(1-(x-3)/abs(x-3))/2((x+3)/3) [-2.955, 7.045, -0.5, 4.5]}