#### Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 39 Subquestion (i) Maths Textbook Solution.

Continuous

Hint:

For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Given:

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

Solution:

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

We have

(LHL at $x=0$ )

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

(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,

(LHL at $x=1$ )

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

(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

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