Examine the continuity of function at x=2 and x =3. f(x)=|x-2|+|x… - Vidyadip

Question

Examine the continuity of function at $x=2$ and $x =3$.$
f(x)=|x-2|+|x-3|
$

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Answer

Given function $f(x)=|x-2|+|x-3|$,
It is written as : $\begin{array}{l} \quad f(x)=\left\{\begin{array}{cc} -(x-2)-(x-3), & 0 \leq x<2 \\ x-2-(x-3), & 2 \leq x<3 \\ x-2+x-3, & x \geq 3 \end{array}\right. \\ \Rightarrow \quad f(x)=\left\{\begin{array}{cl} 5-2 x, & \text { if } 0 \leq x>2 \\ 1, & \text { if } 2 \leq x<3 \\ 2 x-5, & \text { if } x \geq 3 \end{array}\right. \end{array}$
(i) Examining the continuity at $x=2$.
$\begin{array}{l} \text { R.H.L. }=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} 1=1 \\ \begin{aligned} \text { L.H.L. } & =\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}[5-2(2-h)] \\ & =\lim _{h \rightarrow 0}(5-4+2 h)=1\end{aligned}\end{array}$
at $x=2$
$\begin{array}{ll} & f(x)=5-2 x \\ \therefore & f(2)=5-2 \times 2=5-4=1\end{array}$
hence at $x=2$
$f(2)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} f(2-h)=1$
hence function is continuous at $x=2$.
(ii) Examining the continuity at $x=3$
R.H.L.
$\begin{aligned}
\lim _{h \rightarrow 0} f(3+h) & =\lim _{h \rightarrow 0}[2(3+h)-5] \\
& =\lim _{h \rightarrow 0}(6+2 h-5)=1
\end{aligned}$
L.H.L.
$\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0} 1=1$
at $x=3$
$\begin{array}{ll}
& f(x)=2 x-5, \\
\therefore & f(3)=2 \times 3-5=6-5=1 \\
\therefore & f(3)=\lim _{h \rightarrow 0} f(3+h)=\lim _{h \rightarrow 0} f(3-h)=1
\end{array}$
Hence function is continuous at $x=3$.

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