Question

If $f\left(x\right)=\{\begin{array}{cc}x+1,& \text{if}x>0\\ x-1,& \text{if}x\le 0\end{array}$ show that ${\mathrm{li}}_{x\to 0}f\left(x\right)$ does not exists.

Answer 1

For $\underset{x\to 0}{lim}f\left(x\right)$ to exist

$lhl=rhl=$functional value

$L.H.L$

$\underset{x\to 0-h}{lim}\left(f\right(x\left)\right)=\underset{h\to 0}{lim}\left(\right(0-h)-1)=-1$

$\downarrow$

$f\left(x\right)=x-1$

$R.H.L$

$\underset{x\to 0+h}{lim}f\left(x\right)=\underset{h\to 0}{lim}(0+h+1)=1$

$\downarrow$

$x+1$

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

$\therefore$ limit doesn't exist.