f (x) = |x| + |x – 1| at x = 1

Checking the right hand and left hand limits for the given function, we have

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

$=|1-0|+|1-0-1|=1+0=1$

$\lim _{x \rightarrow 1} f(x)=|x|+|x-1|$

$=\lim _{h \rightarrow 0}|1+h|+|1+h-1|=1+0=1$

$\lim _{x \rightarrow 1} f(x)=|x|+|x-1|=|1|+|1-1|=1+0=1$

Now, as

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

Thus, f(x) is continuous at x = 1.