lim_{x→ 0} √(2) - √(1 + cos x) / sin^2 x

Correct option is B. 42
lim_{x→0} \frac{(√(2) - √(1 + cos x))}{sin^2 x}
= lim_{x→0} \frac{(√(2) - √(1 + cos x))}{sin^2 x} \times \frac{(√(2) + √(1 + cos x))}{(√(2) + √(1 + cos x))}
= lim_{x→0} \frac{(2 - (1 + cos x))}{sin^2 x} \times \frac{1}{(√(2) + √(1 + cos x))}
= lim_{x→0} \frac{(1 - cos x)}{sin^2 x} \times \frac{1}{(√(2) + √(1 + cos x))}
= lim_{x→0} \frac{2 sin^2(x/2)}{4 sin^2(x/2)cos^2(x/2)} \times \frac{1}{(√(2) + √(1 + cos x))}
= lim_{x→0} \frac{1}{2 cos^2(x/2)} \times \frac{1}{(√(2) + √(1 + cos x))}
= \frac{1}{2(1)} \times \frac{1}{(√(2) + √(2))} = \frac{1}{4√(2)}
= \frac{√(2)}{8}
This is not matching any of the options. There might be a mistake in the options provided or the question itself.