Question Number 107975 by john santu last updated on 13/Aug/20
$$\:\:\:\:\:\:\frac{\clubsuit{JS}\heartsuit}{\bullet\equiv\bullet} \\ $$$$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right)−\mathrm{tan}\:\left(\mathrm{sin}\:{x}\right)}{{x}−\mathrm{sin}\:{x}\:} \\ $$$$\:\left(\mathrm{2}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} \sqrt{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right)\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right)\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right)\sqrt{...}}}}\:?\: \\ $$$$ \\ $$
Commented by malwaan last updated on 14/Aug/20
$${fantastic}\:{question}\:{and}\:{solutions} \\ $$$${thank}\:{you} \\ $$
Answered by bemath last updated on 13/Aug/20
$$\:\frac{\bullet\mathcal{B}{e}\mathcal{M}{ath}\bullet}{{joss}} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}^{\mathrm{2}} \sqrt{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right)\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right)\sqrt{...}}}\:=\:{L} \\ $$$${let}\:\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right)\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right)\sqrt{...}}}\:=\:{r} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right).{r}\:=\:{r}^{\mathrm{2}} \: \\ $$$$\Rightarrow\:{r}\:=\:\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\:.\:{Then}\: \\ $$$${L}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right) \\ $$$${let}\:\frac{\mathrm{1}}{{x}}\:=\:{q}\:\Rightarrow{L}=\underset{{q}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{q}}{{q}^{\mathrm{2}} } \\ $$$${L}=\underset{{q}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sin}\:^{\mathrm{2}} {q}}{{q}^{\mathrm{2}} }\:=\:\mathrm{2} \\ $$
Answered by bemath last updated on 13/Aug/20
$$\:\:\:\frac{\bullet\mathscr{B}{emath}\bullet}{\vdots\vdots} \\ $$$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{tan}\:{x}−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{tan}\:^{\mathrm{3}} {x}\right)−\left(\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}\:^{\mathrm{3}} {x}\right)}{{x}−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)−\left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)}{\left(\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)}\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{−\frac{\mathrm{3}{x}^{\mathrm{3}} }{\mathrm{6}}}{\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}\right)\:=\:−\mathrm{3}\: \\ $$
Answered by Dwaipayan Shikari last updated on 13/Aug/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}\left({tanx}\right)−{tan}\left({sinx}\right)}{{x}−{sinx}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sinx}−{tanx}}{{x}−{sinx}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}−{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}}{{x}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}=−\mathrm{3} \\ $$$$ \\ $$