Question Number 113786 by bemath last updated on 15/Sep/20
Answered by bobhans last updated on 15/Sep/20
$${recall}\:\mathrm{tan}\:\mathrm{6}{x}\:=\:\mathrm{6}{x}+\frac{\left(\mathrm{6}{x}\right)^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}\left(\mathrm{6}{x}\right)^{\mathrm{5}} }{\mathrm{15}}+… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\left(\frac{\mathrm{4}}{{x}}\right)=\mathrm{1}−\frac{\left(\frac{\mathrm{4}}{{x}}\right)^{\mathrm{2}} }{\mathrm{2}!}+\frac{\left(\frac{\mathrm{4}}{{x}}\right)^{\mathrm{4}} }{\mathrm{4}!}−… \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{6}{x}+\frac{\mathrm{6}^{\mathrm{3}} {x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}.\mathrm{6}^{\mathrm{5}} {x}^{\mathrm{5}} }{\mathrm{15}}+…\right)}{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{8}}{{x}^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{3}{x}^{\mathrm{4}} }−…\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{6}{x}+\frac{\mathrm{6}^{\mathrm{3}} {x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}.\mathrm{6}^{\mathrm{5}} {x}^{\mathrm{5}} }{\mathrm{15}}+…\right)}{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\mathrm{8}+\frac{\mathrm{8}}{\mathrm{3}{x}^{\mathrm{2}} }−…\right)}= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{x}^{\mathrm{2}} \left(\mathrm{6}{x}+\frac{\mathrm{6}^{\mathrm{3}} {x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}.\mathrm{6}^{\mathrm{5}} {x}^{\mathrm{5}} }{\mathrm{15}}+…\right)}{\mathrm{8}+\frac{\mathrm{8}}{\mathrm{3}{x}^{\mathrm{2}} }−…}\:=\mathrm{0} \\ $$
Answered by bemath last updated on 15/Sep/20
