Question

Find

$\underset{x\to 1}{\text{lim}}$ f(x) where f(x)= $\{\begin{array}{cc}{x}^2-1,& x\le 1\\ -x^2-1& x>1\end{array}$

Answer 1

Here f(x) = $\{\begin{array}{cc}{x}^2-1,& x\le 1\\ -x^2-1,& x>1\end{array}$

L.H.L. = $\underset{x\to 1^{-}}{\text{lim}}f\left(x\right)=\underset{x\to 1^{-}}{\text{lim}}(x^2-1)$

Put x=1 - h as $x\to 1,h\to 0$

$\therefore \underset{h\to 0}{\text{lim}}\left[\right(1+h{)}^2-1]=\underset{h\to 0}{\text{lim}}[1+h^2-2h-1]$

= $(0{)}^2-2\times 0=0$

R.H.L. = $\underset{x\to 1^{+}}{\text{lim}}f\left(x\right)=\underset{x\to 1^{+}}{\text{lim}}(-x^2-1)$

Put x=1 - h as x $\to 1,h\to 0$

$\therefore \underset{h\to 0}{\text{lim}}[-(1+h{)}^2-1]=\underset{h\to 0}{\text{lim}}[1-h^2-2h-1]$

=- $(0{)}^2-2\times 0-2=-2$

$\therefore$ L.H.L. $\ne$ R.H.L.

This limit does not exist at x = 1.