Question

If \( f'(x) = 4x\cos^2(x) \sin\left(\frac{x}{4}\right) \), then \( \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} \) is equal to:

• A. \( 4\pi \)
• B. \( \sqrt{2}\pi \)
• C. \( 2\pi \)
• D. \( 2\sqrt{2}\pi \)
• E. \( 0 \)

Hint:

The derivative evaluated at a point directly gives the rate of change at that point, crucial for understanding instantaneous changes in functions modeled by derivatives.

Answer 1

The Correct Option is D

The expression \( \frac{f(\pi + x) - f(\pi)}{x} \) is the definition of the derivative at \( \pi \), which means: \[ \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} = f'(\pi) \] Calculate \( f'(\pi) \) using the given \( f'(x) \): \[ f'(\pi) = 4\pi \cos^2(\pi) \sin\left(\frac{\pi}{4}\right) \] \[ = 4\pi (-1)^2 \sin\left(\frac{\pi}{4}\right) \] \[ = 4\pi \sin\left(\frac{\pi}{4}\right) \] \[ = 4\pi \frac{\sqrt{2}}{2} \] \[ = 2\sqrt{2}\pi \] Thus, the value of the limit is \( 2\sqrt{2}\pi \), matching option (D).