Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{x^3-2x^2-15x+36}{(x-3{)}^2},& x\ne 3\\ p,& x=3\end{array}$ is continuous at $x=3$, then the value of $p$ is

Answer 1

Since $f\left(x\right)$ is continuous at $x=3$, so we have $f\left(3\right)=\underset{x\to 3}{lim}f\left(x\right)$
$\Rightarrow p=\underset{x\to 3}{lim}\frac{x^3-2x^2-15x+36}{(x-3{)}^2}$
$=\underset{x\to 3}{lim}\frac{(x-3{)}^2(x+4)}{(x-3{)}^2}$
$=\underset{x\to 3}{lim}(x+4)=3+4=7$