The function $f(x)=|x|+|x-1|$ is

$\begin{array}{l}f(x)=|x|+|x-1| \\ |x|=\left\{\begin{array}{cc}x \quad x \geqslant 0 \\ -x \quad x<0\end{array}\right. \\ \begin{aligned} f(x) & =|x|+|x-1| \\ & =x+x-1=2 x-1 \\ & =x-x+1=1 \\ & =-x-x+1=-2 x+1\end{aligned}\end{array}$ $|x-1|=\left\{\begin{array}{ll} x-1 & x \geqslant 1 \\ -(x-1) & x<1 \end{array}\right.$

if $x \geqslant 1$

if $0 \leq x<1$

if $\quad x<0$

$\begin{aligned} \therefore \quad L H L=R H L=f(0) \therefore & \text { continous } \\ & \text { at } x=0 .\end{aligned}$

$\begin{array}{l}\therefore \text { at } x=0 \Rightarrow \\ \text { LHL }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(-2 x+1)=1 \\ \text { RHL }=\lim _{x \rightarrow 0^{+}} f(x)=1 \\ f(0)=1\end{array}$

$\begin{array}{l}\text { at } x=1 \Rightarrow \\ L H L=\lim _{x \rightarrow 1^{-}} f(x)=1 \\ R H L=\lim _{x \rightarrow 1^{+}} f(x)=1 \\ f(1)=1 \quad \therefore L H L=R H L=f(1) \therefore \text { continous } \\ \text { at } x=1\end{array}$