Question

\[ \lim_{x \rightarrow \frac{2}{3}} \frac{\sin\left(\pi \cos^2(3x-2)\right)} {9x^2-12x+4} = \ ? \] 

• A. \(\pi \)
• B. \(2\pi \)
• C. \(\frac{\pi}{2} \)
• D. \(\frac{3\pi}{2} \) \bigskip

Hint:

For limits of type \(\frac{\sin f(x)}{g(x)}\), convert to small-angle form \(\sin t \sim t\).

Answer 1

The Correct Option is A

Concept: Use expansion of \(\sin t \sim t\) for small \(t\) and factorization of quadratic expressions.

Step 1: Factor denominator.
\[ 9x^{2}-12x+4=(3x-2)^{2} \]

Step 2: Approximate numerator.
As \(x \to \frac{2}{3}\), \(3x-2 \to 0\). Also, \[ \cos^{2}(3x-2)\approx 1-(3x-2)^{2} \] \[ \sin(\pi\cos^{2}(3x-2)) \approx \sin(\pi - \pi(3x-2)^2) \] \[ = \sin(\pi(3x-2)^2) \approx \pi(3x-2)^2 \]

Step 3: Evaluate limit.
\[ \frac{\pi(3x-2)^2}{(3x-2)^2}=\pi \]