Discuss the continuity of the f(x) at the indicated points:

Solution:

(i) Given: $f(x)=|x|+|x-1|$

We have

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0}[|0-h|+|0-h-1|]=1$

$(\mathrm{RHL}$ at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0}[|0+h|+|0+h-1|]=1$

Also, $f(0)=|0|+|0-1|=0+1=1$

Now,

$(\mathrm{LHL}$ at $x=1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}(|1-h|+|1-h-1|)=1+0=1$

$(\mathrm{RHL}$ at $x=1)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}(|1+h|+|1+h-1|)=1+0=1$

Also, $f(1)=|1|+|1-1|=1+0=1$

$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0) \quad$ and $\quad \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$

Hence, $f(x)$ is continuous at $x=0,1$.

(ii) Given: $f(x)=|x-1|+|x+1|$

We have

$(\mathrm{LHL}$ at $x=-1)=\lim _{x \rightarrow-1^{-}} f(x)=\lim _{h \rightarrow 0} f(-1-h)=\lim _{h \rightarrow 0}[|-1-h-1|+|-1-h+1|]=2+0=2$

$(\mathrm{RHL}$ at $x=-1)=\lim _{x \rightarrow-1^{+}} f(x)=\lim _{h \rightarrow 0} f(-1+h)=\lim _{h \rightarrow 0}[|-1+h-1|+|-1+h+1|]=2+0=2$

Also, $f(-1)=|-1-1|+|-1+1|=|-2|=2$

Now,

$(\mathrm{LHL}$ at $x=1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}(|1-h-1|+|1-h+1|)=0+2=2$

(RHL at $x=1)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}(|1+h-1|+|1+h+1|)=0+2=2$

Also, $f(1)=|1+1|+|1-1|=2$

$\therefore \lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{+}} f(x)=f(-1) \quad$ and $\quad \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$

Hence, $f(x)$ is continuous at $x=-1,1$.