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