Question

$\underset{x\to 0}{\lim }\frac{2\sin x^{\circ }-\sin 2x^{\circ }}{x^3}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{2\sin x°-\sin \left(2x°\right)}{x^3}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2\sin \left({\displaystyle \frac{\mathrm{\pi }x}{180}}\right)-\sin \left({\displaystyle \frac{2\mathrm{\pi }x}{180}}\right)}{x^3}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2\sin \left({\displaystyle \frac{\mathrm{\pi }x}{180}}\right)-2\sin \left({\displaystyle \frac{\mathrm{\pi }x}{180}}\right)\times \cos \left({\displaystyle \frac{\mathrm{\pi }x}{180}}\right)}{x^3}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2\sin \left({\displaystyle \frac{\mathrm{\pi }x}{180}}\right)\left[1-\cos \left({\displaystyle \frac{\mathrm{\pi }x}{180}}\right)\right]}{x^3}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2\sin \left({\displaystyle \frac{\mathrm{\pi }x}{180}}\right)\times 2{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }x}{360}}\right)}{x\times x^2}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{4\sin \left[{\displaystyle \frac{\mathrm{\pi }x}{180}}\right]\times {\sin }^2\left[{\displaystyle \frac{\mathrm{\pi }x}{360}}\right]}{{\displaystyle \frac{\pi x}{180}}\times {\displaystyle \frac{\mathrm{\pi }x}{360}}\times {\displaystyle \frac{\mathrm{\pi }x}{360}}}\times \frac{\mathrm{\pi }}{180}\times {\left(\frac{\mathrm{\pi }}{360}\right)}^2\right] \\ =4\underset{x\to 0}{\lim }\left[\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }x}{180}}\right)}{{\displaystyle \frac{\mathrm{\pi }x}{180}}}\times \frac{\sin \left({\displaystyle \frac{\mathrm{\pi }x}{360}}\right)\times \sin \left({\displaystyle \frac{\mathrm{\pi }x}{360}}\right)}{{\displaystyle \frac{\mathrm{\pi }x}{360}}\times {\displaystyle \frac{\mathrm{\pi }x}{360}}}\times \frac{{\mathrm{\pi }}^3}{180\times {360}^2}\right] \\ =4\times 1\times 1\times 1\times \frac{{\mathrm{\pi }}^3}{180\times 360\times 360} \\ ={\left(\frac{\mathrm{\pi }}{180}\right)}^3 \end{gathered}$$