Functions Question 581
Question: If $ f(x)= \begin{cases} & {e^{1/x}},\ when\ x\ne 0 \\ & 0,\ \ \ \ \ when\ x=0 \\ \end{cases} . $ , then
Options:
A) $ \underset{x\to 0+}{\mathop{\lim }},f(x)=e $
B) $ \underset{x\to 0+}{\mathop{\lim }},f(x)=0 $
C) $ f(x) $ is discontinuous at $ x=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(0)=0 $ $ \underset{x\to 0-}{\mathop{\lim }},f(x)=\underset{h\to 0}{\mathop{\lim }}{e^{-1/h}}=0 $ and $ \underset{x\to 0+}{\mathop{\lim }},f(x)=\underset{h\to 0}{\mathop{\lim }}{e^{1/h}}=\infty $ Hence function is discontinuous at $ x=0 $ .
BETA
