Question

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

Answer 1

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

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

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

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

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

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

$\Rightarrow k=0$

Thus, the value of k is 0.


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