Question Number 105643 by john santu last updated on 30/Jul/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left({x}−\mathrm{sin}\:{x}\right)}{\mathrm{2}−\sqrt{\mathrm{4}+\mathrm{sin}\:^{\mathrm{3}} \:\mathrm{2}{x}}} \\ $$
Answered by bramlex last updated on 30/Jul/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left({x}−\left({x}−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} \right)\right)}{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+\frac{\mathrm{sin}\:^{\mathrm{3}} \mathrm{2}{x}}{\mathrm{4}}}}\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} \right)}{\mathrm{2}\left(\mathrm{1}−\left(\mathrm{1}+\frac{\mathrm{sin}\:^{\mathrm{3}} \mathrm{2}{x}}{\mathrm{8}}\right)\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} \right)}{−\left(\frac{\mathrm{8}}{\mathrm{4}}{x}^{\mathrm{3}} \right)}\:=\:−\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{2}}\:=\:−\frac{\mathrm{1}}{\mathrm{12}}\blacktriangle \\ $$
Answered by Dwaipayan Shikari last updated on 30/Jul/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{tan}\left({x}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)}{\mathrm{4}−\mathrm{4}−{sin}^{\mathrm{3}} \mathrm{2}{x}}\left(\mathrm{2}+\sqrt{\mathrm{4}+{sin}^{\mathrm{3}} \mathrm{2}{x}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{tan}\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}{−\left(\mathrm{2}{x}\right)^{\mathrm{3}} }\left(\mathrm{2}+\sqrt{\mathrm{4}+\mathrm{8}{x}^{\mathrm{3}} }\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}{−\mathrm{8}{x}^{\mathrm{3}} }\left(\mathrm{2}+\sqrt{\mathrm{4}}\right)=−\frac{\mathrm{1}}{\mathrm{12}}\:\:\:\:\left\{\:\:\:\left({sin}\mathrm{2}{x}\right)^{\mathrm{3}} \rightarrow\mathrm{8}{x}^{\mathrm{3}} \right. \\ $$