Question

Find the value of constant 'k' so that the function f(x) defined as; $f\left(x\right)=\{\begin{array}{cc}\frac{x^2-2x-3}{x+3},& x\ne -1\\ k,& x=-1\end{array}$ is continuous at $x=-1$.

Answer 1

$f\left(x\right)$ is continuous at $x=-1$ means $li{m}_{x\to -1}f\left(x\right)=f(-1)$......$\left(1\right)$

$li{m}_{x\to -1}\frac{x^2-2x-3}{x+3}$ is defined at $x=-1$

$\therefore$ we can find limit directly by putting $x=-1$ in limit.

$li{m}_{x\to -1}\frac{x^2-2x-3}{x+3}=\frac{(-1{)}^2-2(-1)-3}{-1+3}=\frac{1+2-3}{2}=0$

By $\left(1\right),f\left(x\right)=0=k$

So value of $k=0.$