Question

If  f(x) =\(\left\lbrace \begin{matrix}\dfrac{\sin [\rm x]}{[\rm x]}, \ \ [\rm x] \neq 0 \\\ 0, \ \ [\rm x] = 0\end{matrix} \right.\), where [x] is the Greatest Integer but not larger than x, then \(\rm \displaystyle\lim_{x \rightarrow 0} f(x)\) is
1. -1
2. 0
3. 1
4. Does not exist

Answer 1

Correct Answer - Option 4 : Does not exist

Concept:

Greatest Integer Function:

Greatest Integer Function [x] indicates an integral part of the real number x which is a nearest and smaller integer to x. It is also known as floor of x

• In general, If n≤ x ≤ n+1 Then [x] = n     (n ∈ Integer)
• Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

 

Calculation:

Given: 

f(x) =\(\left\lbrace \begin{matrix}\dfrac{\sin [\rm x]}{[\rm x]}, \ \ [\rm x] \neq 0 \\\ 0, \ \ [\rm x] = 0\end{matrix} \right.\)

f(x) =\( \left\lbrace \begin{matrix}\dfrac{\rm \sin (-1)}{-1} = \sin 1, \ \ -1 \leq \rm x < 0 \\\ 0, \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\rm \ \ 0 \leq x <1\end{matrix} \right.\)

\(\rm \displaystyle\lim_{x \rightarrow 0^-} f(x) = \sin 1\)

\(\rm \displaystyle\lim_{x \rightarrow 0^+} f(x) =0\)

\(\rm \displaystyle\lim_{x \rightarrow 0^-} f(x) \ne \rm \displaystyle\lim_{x \rightarrow 0^+} f(x)\)

So, \(\rm \displaystyle\lim_{x \rightarrow 0} f(x)\) doesn't exists