Limits Continuity And Differentiability Question 101
Question: Let $ f(x)= \begin{cases} & {5^{1/x}},x<0 \\ & \lambda [x],x\ge 0 \\ \end{cases} . $ and $ \lambda \in R $ , then at x = 0
Options:
A) f is discontinuous
B) f is continuous only, if $ \lambda =0 $
C) f is continuous only, whatever $ \lambda $ may be
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
As we know. A function f(x) is said to be continuous as a point x = a iff $ \underset{x\to a}{\mathop{\lim }}f(x)=f(a), $ otherwise not continuous. Thus f(x) is continuous at x = a iff $ \underset{x\to {a^{-}}}{\mathop{\lim }}f(x)=\underset{x\to {a^{+}}}{\mathop{\lim }}f(x)=f(a) $ Since, $ f(x)= \begin{matrix} {5^{1/x}}, & x<0 \\ \lambda [x], & x\ge 0 \\ \end{matrix}and\lambda \in R . $ RHL $ atx=0:\underset{x\to {0^{+}}}{\mathop{\lim }}f(x)=\underset{x\to {0^{+}}}{\mathop{\lim }}\lambda [x] $
$ =\underset{h\to 0}{\mathop{\lim }}\lambda [h]=0 $ LHL at $ x=0:\underset{x\to {0^{-}}}{\mathop{\lim }}f(x)=\underset{x\to {0^{-}}}{\mathop{\lim }}{5^{1/x}} $
$ =\underset{h\to 0}{\mathop{\lim }}{5^{-1/h}}={5^{\infty }}=\infty $ and $ f(0)=\lambda [0]=0. $ Since, LHL $ \ne $ RHL
$ \therefore $
$ f(x) $ is not continuous.
BETA
