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

Function f is continuous at a point a if the following conditions are satisfied.

1. $f\left(a\right)$ is defined

2. ${lim}_{x\to a}f\left(x\right)$ exists

3. ${lim}_{x\to a}f\left(x\right)=f\left(a\right)$

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

so using L'hospital rule....(differentiating on both sides i.e, deno and numer), we get

$2(x+3)=k$

now substitute $x=3$

=> $k=12$