Question

If \([\,]\) denotes the greatest integer function, then \[ \lim_{x\to \frac{\pi}{2}^{+}} \frac{[\sin x]-[\cos x]+1}{2} = \]

• A. \(0\)
• B. \(-\frac{1}{2}\)
• C. \(\frac{1}{2}\)
• D. \(1\)

Hint:

For greatest integer function problems, first determine the interval in which the function value lies. Near \(x=\frac{\pi}{2}^{+}\), \[ 0<\sin x<1 \Rightarrow [\sin x]=0, \] and \[ -1<\cos x<0 \Rightarrow [\cos x]=-1. \]

Answer 1

The Correct Option is D

Step 1: Analyze \(\sin x\) as \(x\to \frac{\pi}{2}^{+}\).
When \[ x\to \frac{\pi}{2}^{+}, \] we have \[ \sin x<1 \] but very close to \(1\).
Therefore, \[ 0<\sin x<1. \] Hence, \[ [\sin x]=0. \]

Step 2: Analyze \(\cos x\) as \(x\to \frac{\pi}{2}^{+}\).
For values of \(x\) just greater than \[ \frac{\pi}{2}, \] \(\cos x\) becomes a small negative number.
Thus, \[ -1<\cos x<0. \] Therefore, \[ [\cos x]=-1. \]

Step 3: Substitute into the expression.
\[ \frac{[\sin x]-[\cos x]+1}{2} = \frac{0-(-1)+1}{2}. \] \[ = \frac{2}{2}. \] \[ =1. \]

Step 4: Evaluate the limit.
Since the expression remains constant in a right neighbourhood of \[ \frac{\pi}{2}, \] the limit is \[ 1. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{1} \]