Question Number 173498 by blackmamba last updated on 12/Jul/22
$$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{8cot}\:\left({x}\right)+\mathrm{9}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}}\right)}{\mathrm{12}\:\mathrm{csc}\:\left({x}\right)−\mathrm{4sin}\:\left(\frac{\mathrm{1}}{{x}}\right)}\:=? \\ $$
Commented by blackmamba last updated on 13/Jul/22
$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{8cos}\:{x}\left(\frac{{x}}{\mathrm{sin}\:{x}}\right)+\mathrm{9}{x}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}}\right)}{\frac{\mathrm{12}{x}}{\mathrm{sin}\:{x}}−\mathrm{4}{x}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$$\:=\:\frac{\mathrm{8}+\mathrm{0}}{\mathrm{12}−\mathrm{0}}\:=\:\frac{\mathrm{8}}{\mathrm{12}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Answered by aleks041103 last updated on 12/Jul/22
$${L}=\left(\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{8}{cot}\left({x}\right)}{\mathrm{12}{csc}\left({x}\right)}\right)\left(\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}+\frac{\mathrm{9}}{\mathrm{8}}{tan}\left({x}\right){sin}\left(\mathrm{1}/{x}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}{sin}\left({x}\right){sin}\left(\mathrm{1}/{x}\right)}\right)= \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}{cos}\left(\mathrm{0}\right).\frac{\mathrm{1}+\frac{\mathrm{9}}{\mathrm{8}}.\mathrm{0}.\left({sth}\:{finite}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}.\mathrm{0}.\left({sth}\:{finite}\right)}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{Ans}.=\frac{\mathrm{2}}{\mathrm{3}} \\ $$