Functions Question 677
Question: If $ f(x)= \begin{cases} & x\sin \frac{1}{x},\ \ \ \ \ x\ne 0 \\ & \ \ \ \ \ \ 0,\ \ \ \ \ x=0 \\ \end{cases} . $ , then $ \underset{x\to 0}{\mathop{\lim }},f(x)= $
[IIT 1988; MNR 1988; SCRA 1996; UPSEAT 2000, 01]
Options:
A) 1
B) 0
C) ?1
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Here $ f(0)=0 $ Since $ -1\le \sin \frac{1}{x}\le 1\Rightarrow -|x|\le x\sin \frac{1}{x}\le |x| $ We know that $ \underset{x\to 0}{\mathop{\lim }},|x|,=0 $ and $ \underset{x\to 0}{\mathop{\lim }},|x|,=0 $ In this way $ \underset{x\to 0}{\mathop{\lim }},f(x)=0. $
BETA
