49. \[ \lim_{w \to -4} \frac{\sin(\pi w)}{w^2 - 16} \]






Answer is: option1

\(-\frac{\pi}{8}\)

Solution:

Substituting \( w = -4 \):

\[ \sin(\pi(-4)) = \sin(-4\pi) = \sin(4\pi) = 0 \]

\[ (-4)^2 - 16 = 16 - 16 = 0 \]

Since we get \( \frac{0}{0} \), we apply **L'Hôpital's Rule**, which states:

\[ \lim_{w \to c} \frac{f(w)}{g(w)} = \lim_{w \to c} \frac{f'(w)}{g'(w)} \]

Numerator:

\[ \frac{d}{dw} \sin(\pi w) = \pi \cos(\pi w) \]

Denominator:

\[ \frac{d}{dw} (w^2 - 16) = 2w \]

Substituting \( w = -4 \):

\[ \cos(\pi(-4)) = \cos(-4\pi) = \cos(4\pi) = 1 \]

\[ 2(-4) = -8 \]

\[ \frac{\pi \cdot 1}{-8} = -\frac{\pi}{8} \]

Final Answer: \( -\frac{\pi}{8} \)

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