Q: $ \large If \; f(x) = \left\{\begin{array}{ll} \frac{sin[x]}{[x]} \; , for \; [x] \ne 0 \\ 0 \; , for \; [x] = 0 \end{array} \right. $ where [x] denotes the greatest integer less than or equal to x, then $\lim_{x \rightarrow 0} f(x)$ equals to
(A) 1
(B) 0
(C) –1
(D) None of these
Sol: $\lim_{x \rightarrow 0^-} f(x)= \lim_{x \rightarrow 0^-} \frac{sin[x]}{[x]}$
$\large = \frac{sin(-1)}{(-1)} = sin1 $
and , $\lim_{x \rightarrow 0^+} f(x)= 0 $ as it is given that f(x) = 0 for [x]=0
So , $\lim_{x \rightarrow 0} f(x) $ does not exist
Hence (D) is the correct answer.