Q.44. \left| \lim_{x \to 0} \frac{\cos(7x^\circ) - \cos(2x^\circ)}{x^2} \right| is
A. -\frac{45}{2}\pi ^{2}
B. -\frac{45}{2}\pi
C. \frac{-\pi ^{2}}{1440}
D. \frac{-\pi ^{2}}{2880}
Answer:- C. \frac{-\pi ^{2}}{1440}
Explanation :-
\lim_{x \to 0} \frac{\cos(7x^\circ) - \cos(2x^\circ)}{x^2} = \lim_{x \to 0} \frac{\cos\left(\frac{7\pi}{180}\right)x - \cos\left(\frac{2\pi}{180}\right)x}{x^2} = \frac{\left(\frac{2\pi}{180}\right)^2 - \left(\frac{7\pi}{180}\right)^2}{2} ...[\cdots] \lim_{x \to 0} \frac{\cos(mx) - \cos(nx)}{x^2} = \frac{n^2 - m^2}{2} = \frac{-\pi^2}{1440}