Question:
Let $f(x) =\left\{ \begin{array}{ll} (x^2 - 16) /(x-4) ,& {\rm if}~ x < 4, \\ cx^2 +8,& {\rm if}~ x \geq4. \end{array} \right. $
Discuss whether there is any value(s) of c such that $f(x)$ is contentious at $x = 4$.
Attempt:
To check the continuity of $f(x)$ at $x = 4$, we need to check if the left limit and right limit of $f(x)$ as $x$ approaches $4$ are equal, and if they are equal to $f(4)$.
We have, \begin{align*} \lim_{x\to4^-} f(x) &= \lim_{x\to4^-} \frac{x^2 - 16}{x - 4} = 8. \end{align*}
On the other hand, we have, \begin{align*} \lim_{x\to4^+} f(x) &= \lim_{x\to4^+} cx^2 + 8 = 16c + 8. \end{align*}
Therefore, in order for $f(x)$ to be continuous at $x=4$, we need to have $8 = 16c+8$, which implies $c=0$. Thus, $f(x)$ is continuous at $x=4$ if and only if $c=0$.