Question

\[\lim_{x \to 0} \frac{2 \sin x^\circ - \sin 2 x^\circ}{x^3}\] 

Answer 1

\[\lim_{x \to 0} \left[ \frac{2 \sin x° - \sin \left( 2x° \right)}{x^3} \right]\] 
\[ = \lim_{x \to 0} \left[ \frac{2 \sin \left( \frac{\pi x}{180} \right) - \sin \left( \frac{2\pi x}{180} \right)}{x^3} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin \left( \frac{\pi x}{180} \right) - 2 \sin \left( \frac{\pi x}{180} \right) \times \cos\left( \frac{\pi x}{180} \right)}{x^3} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin \left( \frac{\pi x}{180} \right) \left[ 1 - \cos \left( \frac{\pi x}{180} \right) \right]}{x^3} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin \left( \frac{\pi x}{180} \right) \times 2 \sin^2 \left( \frac{\pi x}{360} \right)}{x \times x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{4 \sin \left[ \frac{\pi x}{180} \right] \times \sin^2 \left[ \frac{\pi x}{360} \right]}{\frac{\pi x}{180} \times \frac{\pi x}{360} \times \frac{\pi x}{360}} \times \frac{\pi}{180} \times \left( \frac{\pi}{360} \right)^2 \right]\]
\[ = 4 \lim_{x \to 0} \left[ \frac{\sin \left( \frac{\pi x}{180} \right)}{\frac{\pi x}{180}} \times \frac{\sin \left( \frac{\pi x}{360} \right) \times \sin \left( \frac{\pi x}{360} \right)}{\frac{\pi x}{360} \times \frac{\pi x}{360}} \times \frac{\pi^3}{180 \times {360}^2} \right]\]
\[ = 4 \times 1 \times 1 \times 1 \times \frac{\pi^3}{180 \times 360 \times 360}\]
\[ = \left( \frac{\pi}{180} \right)^3\]