Answer
$b=-6$
$\lim\limits_{x \to 2} f(x)=0$
Work Step by Step
We are given the function:
$f(x)=\begin{cases}
3x+b,\text{ if }x\leq 2\\
x-2,\text{ if }x>2
\end{cases}$
For $\lim\limits_{x \to 2}$ to exist we must have:
$\lim\limits_{x \to 2^-} f(x)=\lim\limits_{x \to 2^+} f(x)$ (1)
Determine the left hand limit and the right hand limit:
$\lim\limits_{x \to 2^-} f(x)=\lim\limits_{x \to 2^-} (3x+b)=3(2)+b=6+b$
$\lim\limits_{x \to 2^+} f(x)=\lim\limits_{x \to 2^+} (x-2)=2-2=0$
For the condition (1) to be checked, we must have:
$6+b=0\Rightarrow b=-6$
The value of the limit is:
$\lim\limits_{x \to 2} f(x)=0$
