Answer
Continuous
Work Step by Step
We are given the function:
$f(x)=\begin{cases}
\dfrac{x^2-4x+3}{x-3},\text{ if }x\not=3\\
2,\text{ if }x=3
\end{cases}$
We use the continuity checklist to determine if $f$ is continuous in $a=3$:
1) $f$ is defined for $a=3$.
2) $\lim\limits_{x \to 3^-} f(x)=\lim\limits_{x \to 3^-} \dfrac{x^2-4x+3}{x-3}=\lim\limits_{x \to 3^-} \dfrac{(x-3)(x-1)}{x-3}$
$=\lim\limits_{x \to 3^-} (x-1)=3-1=2$
$\lim\limits_{x \to 3^+} f(x)=\lim\limits_{x \to 3^+} \dfrac{x^2-4x+3}{x-3}=\lim\limits_{x \to 3^+} \dfrac{(x-3)(x-1)}{x-3}$
$=\lim\limits_{x \to 3^+} (x-1)=3-1=2$
$\lim\limits_{x \to 3^-} f(x)=\lim\limits_{x \to 3^+} f(x)$
Therefore $\lim\limits_{x \to 3} f(x)$ exists.
3) $f(3)=2$
$\lim\limits_{x \to 3}=2$
So $f(1)=\lim\limits_{x \to 3} f(x)$.
Conditions 1, 2, 3 are satisfied, therefore the function is continuous in $a=3$.
