Question

If f(x) = $\{\begin{array}{cc}|x|+1,& x<0\\ 0,& x=0\\ |x|-1,& x>0\end{array}$
for what values of a does $\underset{x\to a}{\text{lim}}$ f(x) exist?

Answer 1

Here f(x) = $\{\begin{array}{cc}|x|+1,& x<0\\ 0,& x=0\\ |x|-1,& x>0\end{array}$

$\Rightarrow f\left(x\right)=\{\begin{array}{cc}-x+1,& x<0\\ 0,& x=0\\ x-1,& x>0\end{array}$

$\therefore \underset{x\to a}{\text{lim}}$ f(x) exists for all a $\ne$ 0.

Now we see whether $\underset{x\to 0}{\text{lim}}$ exist or not

L.H.L. = $\underset{x\to 0^{-}}{\text{lim}}f\left(x\right)=\underset{x\to 0^{-}}{\text{lim}}$ |x|+1

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

$\therefore \underset{h\to 0}{\text{lim}}$ |0-h| +1 = 1

R.H.L. = $\underset{x\to 0^{+}}{\text{lim}}f\left(x\right)=\underset{x\to 0^{+}}{\text{lim}}$ |x| -1

Put x =0 + h as $x\to x\to 0,h\to 0$

$\therefore \underset{x\to 0}{\text{lim}}|0+h|-1=\underset{h\to 0}{\text{lim}}$ h - 1 =-1

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

Thus $\underset{x\to 0}{\text{lim}}$ f(x) does not exist.