Answer 1

 Given : \[f\left( x \right) = \left| x - 1 \right| + \left| x + 1 \right|\]

We have
(LHL at x = −1) =  \[\lim_{x \to - 1^-} f\left( x \right) = \lim_{h \to 0} f\left( - 1 - h \right)\]

\[= \lim_{h \to 0} \left[ \left| - 1 - h - 1 \right| + \left| - 1 - h + 1 \right| \right] = 2 + 0 = 2\]

(RHL at x = −1) =

\[\lim_{x \to - 1^+} f\left( x \right) = \lim_{h \to 0} f\left( - 1 + h \right)\]

\[= \lim_{h \to 0} \left[ \left| - 1 + h - 1 \right| + \left| - 1 + h + 1 \right| \right] = 2 + 0 = 2\]

Also,

\[f\left( - 1 \right) = \left| - 1 - 1 \right| + \left| - 1 + 1 \right| = \left| - 2 \right| = 2\]

Now,
(LHL at x = 1) = \[\lim_{x \to 1^-} f\left( x \right) = \lim_{h \to 0} f\left( 1 - h \right) = \lim_{h \to 0} \left( \left| 1 - h - 1 \right| + \left| 1 - h + 1 \right| \right) = 0 + 2 = 2\]

(RHL at x =1) =  \[\lim_{x \to 1^+} f\left( x \right) = \lim_{h \to 0} f\left( 1 + h \right) = \lim_{h \to 0} \left( \left| 1 + h - 1 \right| + \left| 1 + h + 1 \right| \right) = 0 + 2 = 2\]

Also,

\[f\left( 1 \right) = \left| 1 + 1 \right| + \left| 1 - 1 \right| = 2\]

∴ ​\[\lim_{x \to - 1^-} f\left( x \right) = \lim_{x \to - 1^+} f\left( x \right) = f\left( - 1 \right) and \lim_{x \to 1^-} f\left( x \right) = \lim_{x \to 1^+} f\left( x \right) = f\left( 1 \right)\]

Hence,

\[f\left( x \right)\]   is continuous at  \[x = - 1, 1\] .