Question Number 197514 by cortano12 last updated on 20/Sep/23
$$\:\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sin}\:\mathrm{x}\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)=? \\ $$
Answered by MM42 last updated on 20/Sep/23
$$\frac{\mathrm{1}}{{x}}={t}\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \:{sin}\frac{\mathrm{1}}{{t}}×{sin}^{−\mathrm{1}} {t}\:=\mathrm{0} \\ $$$${tip}\:\:“\:{if}\:\:\:{x}\rightarrow{a}\:\:;\:\:\mid{f}\mid<{k}\:\&\:{lim}_{{x}\rightarrow{a}} {g}=\mathrm{0}\Rightarrow{lim}_{{x}\rightarrow{a}} \:{f}×{g}=\mathrm{0}\:\:'' \\ $$