Let the function f be defined by the equation f(x) = l 3x ; ; ; ;… - Vidyadip

MCQ

Let the function $f$ be defined by the equation $f(x) = \left\{ \begin{array}{l}3x\;\;\;\;\;\;{\rm{if}}\;0 \le x \le 1\\5 - 3x\;\;{\rm{if}}\;{\rm{1}} < x \le 2\end{array} \right.,$ then

A
$\mathop {\lim }\limits_{x \to 1} f(x) = f(1)$
B
$\mathop {\lim }\limits_{x \to 1} f(x) = 3$
C
$\mathop {\lim }\limits_{x \to 1} f(x) = 2$
✓
$\mathop {\lim }\limits_{x \to 1} f(x)$ does not exist

✓

Answer

Correct option: D.

$\mathop {\lim }\limits_{x \to 1} f(x)$ does not exist

$\text{L.H.L.} = \mathop {\lim }\limits_{x \to 1 - 0} f(x) = \mathop {\lim }\limits_{h \to 0} \,\,(1 - h) $
$= \mathop {\lim }\limits_{h \to 0} \,\,3(1 - h)$
$ = \mathop {\lim }\limits_{h \to 0} \,\,(3 - 3h)$
$= 3 - 3\,.\,0$
$= 3$
$\text{R.H.L.} = \mathop {\lim }\limits_{x \to 1 + 0} f(x)= \mathop {\lim \,\,}\limits_{h \to 0} \,f\,(1 + h)$
$= \mathop {\lim }\limits_{h \to 0} \,\,[5 - 3(1 + h)]$
$ = \mathop {\lim }\limits_{h \to 0} \,\,(2 - 3h) = 2 - 3\,.\,0$
$= 2$
Hence $\mathop {\lim }\limits_{x \to 1} \,\,f(x)$ does not exist.

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