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#### Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 39

$f\left ( x \right )$  is continuous at  $x=-1,1$

Hint:

$f\left ( x \right )$  must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.

Given:

$f(x)=|x-1|+|x+1|$

Solution:

$f(x)=|x-1|+|x+1|$

We have

(LHL at $x=-1$ )

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

(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,

(LHL at $x=1$ )

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

(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

\begin{aligned} &f(1)=|1+1|+|1-1|=2+0=2 \\ &\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{-}} f(x)=f(-1) \text { and } \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=f(1) \end{aligned}

Hence $f\left ( x \right )$ is continuous at $x=-1$ and $x=1$ .