Question

If \([x]\) denotes the greatest integer less than or equal to \(x\), then \[ \lim_{x\to 0^-}\frac{\sin([x])}{[x]} \] is equal to ________.

• A. 1
• B. $\sin 1$
• C. -1
• D. 0
• E. $-\sin 1$

Hint:

For $x \to 0^{-}$, the floor function $[x]$ is a constant value of $-1$.

Answer 1

The Correct Option is B

To solve this problem, we need to calculate the limit: 

\(\lim_{x \to 0^-} \frac{\sin([x])}{[x]}\)

where \([x]\) denotes the greatest integer less than or equal to \(x\). Let's break down the analysis:

1. Since we are approaching 0 from the negative side (\(x \to 0^-\)), any number \(x\) very close to zero but negative will have its greatest integer to be \(-1\). That is, for \( -1 < x < 0\), \([x] = -1\).
2. Thus, the expression becomes: \(\frac{\sin(-1)}{-1}\).
3. We know from the properties of sine that \(\sin(-\theta) = -\sin(\theta)\). Therefore: \(\sin(-1) = -\sin(1)\).
4. Substituting this in our expression, we get: \(\frac{-\sin(1)}{-1} = \sin(1)\).

Hence, the limit is \(\sin(1)\), which is the given answer option.

Therefore, the correct answer is: \(\sin 1\)