# $f(x) = \left\{\begin{array}{ll} \frac{sin[x]}{[x]} \; , for \; [x] \ne 0 \\ 0 \; , for \; [x] = 0 \end{array} \right.$

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.