Solve this

The function $f(x)=\left\{\begin{array}{cc}\frac{x^{2}-9}{x-3}, & x \neq 3 \\ 2 x+k, & x=3\end{array}\right.$ is continuous at $x=3$.

$\therefore f(3)=\lim _{x \rightarrow 3} f(x)$

$\Rightarrow 2 \times 3+k=\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}$

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

$\Rightarrow 6+k=\lim _{x \rightarrow 3}(x+3)$

$\Rightarrow 6+k=3+3=6$

$\Rightarrow k=0$

Thus, the value of k is 0.

If $f(x)=\left\{\begin{array}{cl}\frac{x^{2}-9}{x-3}, & x \neq 3 \\ 2 x+k, & x=3\end{array}\right.$ is continuous at $x=3$, then $k=$ ___0___.