Functions Question 114
Question: If $ f(x)= \begin{cases} & x^{2}\sin \frac{1}{x},\ \ \ \text{when }x\ne 0 \\ & 0,when,x=0 \\ \end{cases} . $ , then
Options:
A) $ f(0+0)=1 $
B) $ f(0-0)=1 $
C) f is continuous at $ x=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \underset{x\to {0^{+}}}{\mathop{\lim }}f(x)=x^{2}\sin \frac{1}{x}, $ but $ -1\le \sin \frac{1}{x}\le 1 $ and $ x\to 0 $ Therefore, $ \underset{x\to {0^{+}}}{\mathop{\lim }}f(x)=0=\underset{x\to {0^{-}}}{\mathop{\lim }}f(x)=f(0) $ Hence $ f(x) $ is continuous at $ x=0. $
BETA
