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 → 0+. This helps in handling the modulus and greatest integer functions effectively.

Step 2: Key Formula or Approach:

1. |x| near 3 behaves as x.
2. [x] (greatest integer function) for values slightly less than an integer I is I - 1.

Step 3: Detailed Explanation:

Let x = 3 - h, where h > 0.

1. [3x - 9] = [3(3 - h) - 9] = [9 - 3h - 9] = [-3h]
Since -3h is slightly less than 0, its greatest integer value is -1.

2. |3 - x| = |3 - (3 - h)| = |h| = h

3. |x| = 3 - h

Numerator becomes:
3 - (3 - h) + sin(h) = h + sin h

Also,
cos(9 - 3(3 - h)) = cos(3h)

Denominator becomes:
h × (-1) = -h

So the limit becomes:

lim (h → 0) [(h + sin h) cos(3h)] / (-h)

= - lim (h → 0) (1 + (sin h / h)) × cos(3h)

Now using standard limits:
sin h / h → 1 and cos(3h) → 1

So,
= -(1 + 1)(1) = -2

Step 4: Final Answer:

The limit is -2.