Question

Determine the value of ${}^{\prime }{k}^{\prime }$ 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

$\text{Given that,}f\left(x\right)\text{is continuous at}x=3$

$\underset{x\to 3}{lim}f\left(x\right)=f\left(3\right)$

$\Rightarrow \underset{x\to 3}{lim}f\left(x\right)=\underset{x\to 3}{lim}\frac{(x+3{)}^2-36}{x-3}=k$

$\Rightarrow \underset{x\to 3}{\text{Lt}}\frac{(x+3{)}^2-(6{)}^2}{x-3}=k$

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

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

$\Rightarrow \underset{x\to 3}{lim}x+9=k$

$\Rightarrow k=12$

Thus, $f\left(x\right)$ is continuous at $x=3$, if $k=12$.