Functions Question 365
Question: If $ f(x)= \begin{cases} & \frac{x}{{e^{1/x}}+1},whenx\ne 0 \\ & 0,\text{when }x=0 \\ \end{cases} . $ , then
Options:
A) $ \underset{x\to 0+}{\mathop{\lim }},f(x)=1 $
B) $ \underset{x\to 0-}{\mathop{\lim }},f(x)=1 $
C) $ f(x) $ is continuous at $ x=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(0)=0 $ ; $ f(0-)=\underset{h\to 0}{\mathop{\lim }}\frac{-h}{{e^{-1/h}}+1}=\underset{h\to 0}{\mathop{\lim }}\frac{-h}{1+\frac{1}{{e^{1/h}}}}=0 $ $ f(0+)=\underset{h\to 0}{\mathop{\lim }}\frac{h}{{e^{1/h}}+1}=0. $
BETA
