For each t`in R`,let[t]be the greatest integer less than or equal to t. Then
`lim_(xto1^+)((1-absx+sinabs(1-x))sin(pi/2[1-x]))/(abs(1-x)[1-x])`

Options

A. equals-1
B. equals 1
C. does not exist
D. equals 0

Answer

To solve the limit

\[ \lim_{x \to 1^+} \frac{(1 - |x| + \sin |1 - x|) \sin\left(\frac{\pi}{2} [1 - x]\right)}{|1 - x| [1 - x]} \]

we will follow these steps:

### Step 1: Substitute \( x = 1 + h \)
Let \( h \to 0^+ \) as \( x \to 1^+ \). Therefore, we have:
\[ 1 - x = 1 - (1 + h) = -h \]

### Step 2: Rewrite the limit in terms of \( h \)
Substituting \( x = 1 + h \) into the limit, we get:
\[ \lim_{h \to 0^+} \frac{(1 - |1 + h| + \sin |1 - (1 + h)|) \sin\left(\frac{\pi}{2} [1 - (1 + h)]\right)}{|1 - (1 + h)| [1 - (1 + h)]} \]

### Step 3: Simplify the expressions
Now we simplify each part:
- Since \( |1 + h| = 1 + h \), we have:
\[ 1 - |1 + h| = 1 - (1 + h) = -h \]
- For \( \sin |1 - (1 + h)| = \sin |-h| = \sin h \).
- For \( [1 - (1 + h)] = [ -h] \), since \( h \to 0^+ \), \( -h \) is negative, so \( [ -h] = -1 \).

### Step 4: Substitute back into the limit
Now substituting these simplifications back into the limit gives:
\[ \lim_{h \to 0^+} \frac{(-h + \sin h) \sin\left(\frac{\pi}{2} \cdot (-1)\right)}{| -h| (-h)} \]

### Step 5: Evaluate \( \sin\left(\frac{\pi}{2} \cdot (-1)\right) \)
We know that:
\[ \sin\left(-\frac{\pi}{2}\right) = -1 \]

### Step 6: Substitute and simplify the limit
Now substituting this back into our limit:
\[ \lim_{h \to 0^+} \frac{(-h + \sin h)(-1)}{h(-h)} = \lim_{h \to 0^+} \frac{h - \sin h}{h^2} \]

### Step 7: Apply L'Hôpital's Rule
Since both the numerator and denominator approach 0 as \( h \to 0 \), we can apply L'Hôpital's Rule:
\[ \lim_{h \to 0^+} \frac{1 - \cos h}{2h} \]

### Step 8: Evaluate the limit
As \( h \to 0 \), \( \cos h \to 1 \), thus:
\[ \lim_{h \to 0^+} \frac{1 - 1}{2h} = \lim_{h \to 0^+} \frac{0}{2h} = 0 \]

### Final Answer
Thus, the limit evaluates to:
\[ \boxed{0} \]