Functions Question 297
Question: If $ f(x)= \begin{cases} & \frac{\sin
[x]}{[x]},\text{ when }[x]\ne 0 \\ & ,0,\text{ when }[x]=0 \\ \end{cases} . $ where [x] is greatest integer function, then $ \underset{x\to 0}{\mathop{\lim }},f(x)= $ [IIT 1985; RPET 1995]
Options:
A) ?1
B) 1
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
In closed interval of x = 0 at right hand side [x] = 0 and at left hand side $ [x]=-1. $ Also [0]=0. Therefore function is defined as $ f(x)= \begin{cases} & \frac{\sin ,[x]}{[x]}(-1\le x<0) \\ & \ \ \ \ \ 0\ \ (0\le x<1) \\ \end{cases} . $ \Left hand limit $ =\underset{x\to 0-}{\mathop{\lim }},f(x)=\underset{x\to 0-}{\mathop{\lim }}\frac{\sin ,[x]}{[x]} $ $ =\frac{\sin ,(-1)}{-1}=\sin 1^{c} $ Right hand limit = 0. Hence limit doesn’t exist.
BETA
