Differentiation Question 141
Question: If $ f(x)= \begin{matrix} \frac{{{[x]}^{2}}+\sin [x]}{[x]}for[x]\ne 0 \\ 0for[x]=0 \\ \end{matrix} . $ , where [x] denotes the greatest integer less than or equal to $ x, $ Then $ \underset{x\to 0}{\mathop{\lim }}f(x) $ equals
Options:
A) 1
B) 0
C) -1
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
[d] As $ x\to 0- $ (i.e., approaches 0 from the left), $ [x]=-1. $
$ \therefore \underset{x\to {0^{-}}}{\mathop{\lim }}f(x)=\underset{x\to {0^{-}}}{\mathop{\lim }}\frac{1+\sin (-1)}{-1}=-1+\sin 1 $ Whereas, if $ x\to {0^{+}} $ we get $ [x]=0. $
$ \therefore f(x)=0\Rightarrow \underset{x\to {0^{+}}}{\mathop{\lim }}f(x)=0 $ Thus, $ \underset{x\to 0}{\mathop{\lim }}f(x) $ does not exist.
BETA
