Functions Question 206
Question: Let the function f be defined by the equation $ f(x)= \begin{cases} & 3x \ \ if \ 0\le x\le 1 \\ & 5-3x \ \ if\ 1\lt x\le 2 \ \end{cases} , $ then
[SCRA 1996]
Options:
A) $ \underset{x\to 1}{\mathop{\lim }} f(x)=f(1) $
B) $ \underset{x\to 1}{\mathop{\lim }} f(x)=3 $
C) $ \underset{x\to 1}{\mathop{\lim }} f(x)=2 $
D) $ \underset{x\to 1}{\mathop{\lim }} f(x) $ does not exist
Show Answer
Answer:
Correct Answer: D
Solution:
L.H.L. $ =\underset{x\to 1-0}{\mathop{\lim }} f(x)=\underset{h\to 0}{\mathop{\lim }} (1-h)=\underset{h\to 0}{\mathop{\lim }} 3(1-h) $ $ =\underset{h\to 0}{\mathop{\lim }} (3-3h)=3-3 . 0=3 $ R.H.L. $ =\underset{x\to 1+0}{\mathop{\lim }} f(x)=\underset{h\to 0}{\mathop{\lim }}f (1+h)=\underset{h\to 0}{\mathop{\lim }} [5-3(1+h)] $ $ =\underset{h\to 0}{\mathop{\lim }} (2-3h)=2-3 .,0=2 $ Hence $ \underset{x\to 1}{\mathop{\lim }},f(x) $ does not exist.
BETA
