Question Number 114079 by bemath last updated on 17/Sep/20
Answered by john santu last updated on 17/Sep/20
$${setting}\:\frac{\mathrm{1}}{{x}}\:=\:{m}\:\wedge{m}\rightarrow\mathrm{0} \\ $$$$\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{m}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{4}{m}}{\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{m}\right)\mathrm{sin}\:{m}}\:= \\ $$$$\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{m}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{4}{m}}{\mathrm{2sin}\:^{\mathrm{2}} {m}.\:\mathrm{sin}\:{m}}\:=\:\frac{\mathrm{4}}{\mathrm{2}}\:=\:\mathrm{2} \\ $$
Answered by Olaf last updated on 17/Sep/20
$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\frac{\mathrm{4}}{{x}}}{\left(\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}}{{x}}\right)^{\mathrm{2}} \right)\right){x}^{\mathrm{2}} \frac{\mathrm{1}}{{x}}} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\frac{\mathrm{4}}{{x}}}{\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}}{{x}}\right)^{\mathrm{2}} \right){x}^{\mathrm{2}} \frac{\mathrm{1}}{{x}}} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\frac{\mathrm{4}}{{x}}}{\left(\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right){x}}\:=\:\frac{\mathrm{4}}{\mathrm{2}}\:=\:\mathrm{2} \\ $$$$ \\ $$