Question

Determine the value of 'k' for which the following function is continuous at $x=3,$

$f\left(x\right)=\{\begin{array}{cc}\frac{(x+3{)}^2-36}{x-3},& x\ne 3\\ k,& x=3\end{array}$

Answer 1

Since, the function $f\left(x\right)$ is continuous at $x=3$.

Therefore,

$k={lim}_{x\to 3}\frac{(x+3{)}^2-36}{x-3}$

$k={lim}_{x\to 3}\frac{(x+3{)}^2-6^2}{x-3}$

$k={lim}_{x\to 3}\frac{(x+3-6)(x+3+6)}{x-3}$

$k={lim}_{x\to 3}\frac{(x-3)(x+9)}{x-3}$

$k={lim}_{x\to 3}(x+9)$

$k=3+9=12$

Hence, this is the answer.