Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find which of the functions is continuous or discontinuous at the indicated points:f(x) equals |x minus 4|

Find which of the functions is continuous or discontinuous at the indicated points:

f(x)=\left\{\begin{array}{cl} \frac{|x-4|}{2(x-4)}, \text { if } & x \neq 4 \\ 0, & \text { if } x=4 \end{array}\right.

at x = 4

Answers (1)

Given,

f(x)=\left\{\begin{array}{cl} \frac{|x-4|}{2(x-4)}, \text { if } & x \neq 4 \\ 0, & \text { if } x=4 \end{array}\right. ...(1)

We need to check its continuity at x = 4
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 = 4 if -

\lim _{h \rightarrow 0} f(4-h)=\lim _{h \rightarrow 0} f(4+h)=f(4)

Clearly,

\begin{aligned} &\lim _{\mathrm{h} \mathrm{HL}} \mathrm{f}(4-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{|4-\mathrm{h}-4|}{2(4-\mathrm{h}-4)}\{\text { using equation } 1\}\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{|-\mathrm{h}|}{-2 \mathrm{~h}}\\ &\because \mathrm{h}>0 \text { as defined above. }\\ &\therefore|-h|=h\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h}}{-2 \mathrm{~h}}=-\frac{1}{2} \lim _{\mathrm{h} \rightarrow 0} 1\\ &\therefore \mathrm{LHL}=-1 / 2 \ldots(2) \end{aligned}

Similarly, we proceed for RHL-

\begin{aligned} &{\mathrm{RHL}}=\lim _{\mathrm{h}\rightarrow 0}{\mathrm{f}}(4+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{|4+\mathrm{h}-4|}{2(4+\mathrm{h}-4)}\{\text { using equation } 1\}\\ &\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{|\mathrm{h}|}{2(\mathrm{~h})}\\ &\because \mathrm{h}>0 \text { as defined above. }\\ &\therefore|h|=h \end{aligned}

\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h}}{2 \mathrm{~h}}=\frac{1}{2} \lim _{\mathrm{h} \rightarrow 0} 1

\therefore \mathrm{RHL}=1 / 2 \ldots(3) \\And, f(4)=0\{ \text{using equation 1}\} \ldots(4) \\ \text{From equation 2,3 and 4 we can conclude that} \\ \lim _{h \rightarrow 0} f(4-h) \neq \lim _{h \rightarrow 0} f(4+h) \neq f(4)

∴ f(x) is discontinuous at x = 4

 

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question