calculate-lim-x-0-sin-2sinx-x-2-

Question Number 147685 by mathmax by abdo last updated on 22/Jul/21

calculate \lim_{x \to 0} \frac{\sin (2 sinx)}{x^{2}}

Answered by liberty last updated on 23/Jul/21

\lim_{x \to 0} \frac{\sin (2 \sin x)}{x^{2}} =

\lim_{x \to 0} \frac{2 \sin x - \frac{8 \sin^{3} x}{6}}{x^{2}} =

\frac{1}{6} \lim_{x \to 0} \frac{12 \sin x - 8 \sin^{3} x}{x^{2}} =

\frac{1}{6} \lim_{x \to 0} \frac{\sin x (12 - 8 \sin^{2} x)}{x^{2}} =

\frac{1}{6} \lim_{x \to 0} \frac{12 - 8 \sin^{2} x}{x} = \infty

Answered by mathmax by abdo last updated on 23/Jul/21

sinx \sim x - \frac{x^{3}}{6} \Rightarrow 2 sinx \sim 2 x - \frac{x^{3}}{3} \Rightarrow \sin (2 sinx) \sim \sin (2 x - \frac{x^{3}}{3})

\sim 2 x - \frac{x^{3}}{3} - \frac{1}{6} {(2 x - \frac{x^{3}}{3})}^{3} \Rightarrow \frac{\sin (2 sinx)}{x^{2}} \sim \frac{2}{x} - \frac{x^{2}}{3} - \frac{x}{6} {(2 - \frac{x^{2}}{3})}^{3} \Rightarrow

\lim_{x \to 0} \frac{\sin (2 sinx)}{x^{2}} = \bar{+} \infty