The value of the constant λ, so that the function given below is cont…

$f(x)=\left\{\begin{array}{cc} \frac{x^2-2 x-3}{x+1} & x \neq-1 \\ \lambda & x=-1 \end{array}\right.$

for $f(x)$ to be continous at $x=-1$

$\lim\limits_{x \rightarrow-1} f(x)=f(-1)$

$\Rightarrow \lim\limits_{x \rightarrow-1} \frac{\left(x^2-2 x-3\right)}{x+1}=\lambda$

factorising numerator

$\lim\limits_{x \rightarrow-1} \frac{x^2-3 x+x-3}{x+1}=\lambda$

$\Rightarrow \lim\limits_{x \rightarrow-1} \frac{x(x-3)+1(x-3)}{(x+1)}=\lambda$

$\Rightarrow \lim\limits_{x \rightarrow-1} \frac{(x-3)(x+1)}{(x+1)}=\lambda$

$\Rightarrow \lim\limits_{x \rightarrow-1} x - 3 = \lambda$

$\Rightarrow -1-3=\lambda$

$\Rightarrow \lambda=-4$