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  • Find which of the functions is continuous or discontinuous at the indicated points: f(x) = |x| + |x - 1| at x = 1

Find which of the functions is continuous or discontinuous at the indicated points:
f(x) = |x| + |x - 1| at x = 1

Answers (1)

Given,

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

We need to check its continuity at x = 1
A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-

\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)

Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 1 if -

\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} f(1+h)=f(1)

Then,

\begin{aligned} &\left.\lim _{\mathrm{h} \mathrm{HL}} \mathrm{f}(1-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\{|1-\mathrm{h}|+|1-\mathrm{h}-1|\}_{\{\text {using }} \operatorname{egn} 1\right\}\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0}\{|1-\mathrm{h}|+|-\mathrm{h}|\}\\ &\because \mathrm{h}>0 \text { as defined above and } \mathrm{h} \rightarrow 0\\ &\therefore|-h|=h\\ &\text { And }(1-h)>0 \end{aligned}

\begin{array}{l} \therefore|1-h|=1-h \\ \Rightarrow \quad L H L=h \rightarrow 0\{(1-h)+h\}=\lim _{h \rightarrow 0} 1 \\ \therefore L H L=1 \ldots(2) \end{array}

Similarly, we proceed for RHL-

\begin{aligned} &\lim _{\mathrm{RHL}} \mathrm{h} \rightarrow 0^{\mathrm{f}}(1+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\{|1+\mathrm{h}|+|1+\mathrm{h}-1|\}_{\{\text {using eqn } 1\}}\\ &\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}\{|1+\mathrm{h}|+|\mathrm{h}|\}\\ &\because \mathrm{h}>0 \text { as defined above and } \mathrm{h} \rightarrow 0\\ &\therefore|h|=h \end{aligned}

\begin{aligned} &\text { And }(1+h)>0\\ &\therefore|1+h|=1+h\\ &\Rightarrow \lim _{h \rightarrow 0}\{(1+h)+h\}=\lim _{h \rightarrow 0}(1+2 h)\\ &\therefore \mathrm{RHL}=1+2(0)=1 \ldots(3)\\ &\text { And, }\\ &f(1)=|1|+|1-1|=1\{\text { using egn } 1\} \ldots(4) \end{aligned}

From equation 2,3 and 4 we can conclude that
\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} f(1+h)=f(1)=1
\therefore F(x)  is continuous at  x=1 

Posted by

infoexpert22

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