Functions Question 305
Question: The function $ f(x)=x^{2}\sin \frac{1}{x},,x\ne ,0,f(0),=0 $ at $ x=0 $
[MP PET 2003]
Options:
A) Is continuous but not differentiable
B) Is discontinuous
C) Is having continuous derivative
D) Is continuous and differentiable
Show Answer
Answer:
Correct Answer: D
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 $ \ $ \underset{x\to {0^{+}}}{\mathop{\lim }},f(x)=0=\underset{x\to {0^{-}}}{\mathop{\lim }},f(x)=f(0) $ Therefore $ f(x) $ is continuous at $ x=0 $ . Also, the function $ f(x)=x^{2}\sin \frac{1}{x} $ is differentiable because $ R{f}’(x)=\underset{h\to 0}{\mathop{\lim }},\frac{h^{2}\sin \frac{1}{h}-0}{h}=0 $ , $ L{f}’(x)=\underset{h\to 0}{\mathop{\lim }},\frac{h^{2}\sin (1/-h)}{-h}=0 $ .
BETA
