Question

If [x] is the greatest integer function then \( \lim_{x \to 3^-} \frac{(3-|x| + \sin|3-x|) \cos(9-3x)}{|3-x|[3x-9]} = \)

• A. 0
• B. 1
• C. 2
• D. -2

Hint:

Always substitute \( x = a-h \) for LHL and \( x = a+h \) for RHL when dealing with greatest integer functions near critical points.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

We evaluate the left-hand limit by substituting \( x = 3-h \) where \( h \to 0^+ \). This handles the modulus and greatest integer functions effectively.
Step 2: Key Formula or Approach:

1. \( |x| \) near 3 is \( x \). 2. \( [x] \) for \( x \) slightly less than an integer \( I \) is \( I-1 \).
Step 3: Detailed Explanation:

Let \( x = 3-h \), \( h \textgreater 0 \). 1. \( [3x-9] = [3(3-h)-9] = [9-3h-9] = [-3h] \). Since \( -3h \) is slightly less than 0, the floor is -1. 2. \( |3-x| = |3-(3-h)| = |h| = h \). 3. \( |x| = 3-h \). Numerator: \( 3 - (3-h) + \sin(h) = h + \sin h \). \( \cos(9-3(3-h)) = \cos(3h) \). Denominator: \( h \times (-1) = -h \). Limit expression: \[ \lim_{h \to 0} \frac{(h + \sin h) \cos(3h)}{-h} = -\lim_{h \to 0} \left( 1 + \frac{\sin h}{h} \right) \cos(3h) \] \[ = -(1 + 1)(1) = -2 \]
Step 4: Final Answer:

The limit is -2.