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Question: Let the function f be defined by the equation \(f(x) = \left\{ \begin{matrix} 3x,\text{if}0 \leq x ...

Let the function f be defined by the equation

f(x)={3x,if0≤x≤15−3x,if1<x≤2f(x) = \left\{ \begin{matrix} 3x,\text{if}0 \leq x \leq 1 \\ 5 - 3x,\text{if}1 < x \leq 2 \end{matrix} \right. , then

A

lim⁡x→1f(x)=f(1)\lim_{x \rightarrow 1}f(x) = f(1)

B

lim⁡x→1f(x)=3\lim_{x \rightarrow 1}f(x) = 3

C

lim⁡x→1f(x)=2\lim_{x \rightarrow 1}f(x) = 2

D

lim⁡x→1f(x)\lim_{x \rightarrow 1}f(x) does not exist

Answer

lim⁡x→1f(x)\lim_{x \rightarrow 1}f(x) does not exist

Explanation

Solution

L.H.L.=lim⁡x→1−0f(x)=lim⁡h→0f(1−h)=lim⁡h→03(1−h)= \lim _ { x \rightarrow 1 - 0 } f ( x ) = \lim _ { h \rightarrow 0 } f ( 1 - h ) = \lim _ { h \rightarrow 0 } 3 ( 1 - h )

=lim⁡h→0(3−3h)=3−3.0=3= \lim_{h \rightarrow 0}(3 - 3h) = 3 - 3.0 = 3

R.H.L.=lim⁡x→1+0f(x)=lim⁡h→0f(1+h)=lim⁡h→0[5−3(1+h)]=lim⁡h→0(2−3h)=2−3.0=2= \lim_{x \rightarrow 1 + 0}f(x) = \lim_{h \rightarrow 0}f(1 + h) = \lim_{h \rightarrow 0}\lbrack 5 - 3(1 + h)\rbrack = \lim_{h \rightarrow 0}(2 - 3h) = 2 - 3.0 = 2

Hence lim⁡x→1f(x)\lim_{x \rightarrow 1}f(x) does not exists.