Question

Evaluate the limit: \[ \lim_{x \to 0} \frac{\sin(\pi \cos^2 x)}{x^2}. \]

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

Hint:

For small \( x \), use standard approximations \( \cos x \approx 1 - x^2/2 \) and \( \sin x \approx x \) to simplify limits.

Answer 1

The Correct Option is B

Step 1: Expanding \( \cos^2 x \) for small \( x \) Using \( \cos x \approx 1 - \frac{x^2}{2} \), we approximate: \[ \cos^2 x \approx 1 - x^2. \] Step 2: Expanding \( \sin(\pi \cos^2 x) \) \[ \sin(\pi \cos^2 x) = \sin(\pi (1 - x^2)) \approx \sin(\pi - \pi x^2). \] Using \( \sin(\pi - \theta) = \sin\theta \), we get: \[ \sin(\pi - \pi x^2) \approx \sin(\pi x^2) \approx \pi x^2. \] Step 3: Evaluating the limit \[ \lim_{x \to 0} \frac{\sin(\pi \cos^2 x)}{x^2} = \lim_{x \to 0} \frac{\pi x^2}{x^2} = \pi. \]

Answer 2

Given the limit:
\[ \lim_{x \to 0} \frac{\sin(\pi \cos^2 x)}{x^2} \]

Step 1: As \( x \to 0 \), \(\cos x \to 1\), so \(\cos^2 x \to 1\).
Hence, inside the sine function:
\[ \pi \cos^2 x \to \pi \times 1 = \pi \] and \(\sin(\pi) = 0\). So the expression is of the form \(\frac{0}{0}\), allowing use of expansions.

Step 2: Use the Taylor expansion for \(\cos x\) near 0:
\[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots \] \[ \cos^2 x = \left(1 - \frac{x^2}{2} + \dots\right)^2 = 1 - x^2 + \dots \]

Step 3: Substitute into \(\sin(\pi \cos^2 x)\):
\[ \sin(\pi \cos^2 x) = \sin\left(\pi \left[1 - x^2 + \dots \right]\right) = \sin(\pi - \pi x^2 + \dots) \]

Step 4: Use the sine identity:
\[ \sin(\pi - \theta) = \sin \theta \] So:
\[ \sin(\pi - \pi x^2 + \dots) = \sin(\pi x^2 + \dots) \]

Step 5: For small \( x \), \(\sin(\pi x^2) \approx \pi x^2\). So:
\[ \sin(\pi \cos^2 x) \approx \pi x^2 \]

Step 6: Substitute back into the limit:
\[ \lim_{x \to 0} \frac{\sin(\pi \cos^2 x)}{x^2} \approx \lim_{x \to 0} \frac{\pi x^2}{x^2} = \pi \]

Therefore,
\[ \boxed{\pi} \]