Solution: As at $x=0$, $f(x)$ is continuous .
Therefore,
Left Hand Limit
=$\lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lambda\left(x^{2}-2 x\right)=\lambda(0-0)=0$
And Right Hand Limit
= $\lim _{x \rightarrow 0^{-}} f(x)=f(0)=4 x+1=4 \times 0+1=1$
Given here, L.H.L. $\neq$ R.H.L.
This says that $0=1$, which is not possible.
Again, $f(x)$ is continuous at $x=1$
As a result,
$\lim _{x \rightarrow 1^{-}} f(x)=f(-1)=\lambda\left(x^{2}-2 x\right)=\lambda(1+2)=3 \lambda$
And $\lim _{x \rightarrow-1} f(x)=f(1)=4 x+1=4 \times 1+1=5$
Now say that, L.H.L. = R.H.L.
$\Rightarrow 3 \lambda=5$
$\Rightarrow \lambda=\frac{5}{3}$