Question

Limit (x --> 0) (sin 2x + sin 6x)/(sin 5x - sin 3x)

Answer 1

$$\begin{gathered} \underset{\mathrm{x}\to 0}{\lim }\frac{\sin 2\mathrm{x}+\sin 6\mathrm{x}}{\sin 5\mathrm{x}-\sin 3\mathrm{x}} \\ =\underset{\mathrm{x}\to 0}{\lim }\frac{{\displaystyle \frac{\sin 2\mathrm{x}}{\mathrm{x}}+\frac{\sin 6\mathrm{x}}{\mathrm{x}}}}{{\displaystyle \frac{\sin 5\mathrm{x}}{\mathrm{x}}-\frac{\sin 3\mathrm{x}}{\mathrm{x}}}} \\ =\underset{\mathrm{x}\to 0}{\lim }\frac{2\left({\displaystyle \frac{\sin 2\mathrm{x}}{2\mathrm{x}}}\right)+6\left({\displaystyle \frac{\sin 6\mathrm{x}}{6\mathrm{x}}}\right)}{5\left({\displaystyle \frac{\sin 5\mathrm{x}}{5\mathrm{x}}}\right)-3\left({\displaystyle \frac{\sin 3\mathrm{x}}{3\mathrm{x}}}\right)} \\ =\frac{2\underset{\mathrm{x}\to 0}{\lim }\left({\displaystyle \frac{\sin 2\mathrm{x}}{2\mathrm{x}}}\right)+6\underset{\mathrm{x}\to 0}{\lim }\left({\displaystyle \frac{\sin 6\mathrm{x}}{\mathrm{x}}}\right)}{5\underset{\mathrm{x}\to 0}{\lim }\left({\displaystyle \frac{\sin 5\mathrm{x}}{5\mathrm{x}}}\right)-3\underset{\mathrm{x}\to 0}{\lim }\left({\displaystyle \frac{\sin 3\mathrm{x}}{3\mathrm{x}}}\right)} \\ =\frac{2.1+6.1}{5.1-3.1} \\ =\frac{8}{2} \\ =4 \end{gathered}$$