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Question Number 157314 by cortano last updated on 22/Oct/21

lim_(x→∞) ((3 ln (1+5tan (4/x)))/(x (1−cos (6/x)))) =?

$$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}\:\mathrm{ln}\:\left(\mathrm{1}+\mathrm{5tan}\:\frac{\mathrm{4}}{{x}}\right)}{{x}\:\left(\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{6}}{{x}}\right)}\:=? \\ $$

Commented by john_santu last updated on 22/Oct/21

let (1/x) = u and u→0 lim_(u→0) ((3u ln (1+5tan 4u))/(1−cos 6u)) = lim_(u→0) ((3u ln (1+5tan 4u))/(2sin^2 3u )) = lim_(u→0) (((3u)/(2sin 3u))).lim_(u→0) (((ln (1+5tan 4u))/(sin 3u))) =(1/2)×lim_(u→0) ((ln (1+5tan 4u))/(5tan 4u)) ×((5tan 4u)/(sin 3u)) =(1/2)×1×((20)/3)=((10)/3)

$$\:{let}\:\frac{\mathrm{1}}{{x}}\:=\:{u}\:{and}\:{u}\rightarrow\mathrm{0} \\ $$$$\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}{u}\:\mathrm{ln}\:\left(\mathrm{1}+\mathrm{5tan}\:\mathrm{4}{u}\right)}{\mathrm{1}−\mathrm{cos}\:\mathrm{6}{u}} \\ $$$$=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}{u}\:\mathrm{ln}\:\left(\mathrm{1}+\mathrm{5tan}\:\mathrm{4}{u}\right)}{\mathrm{2sin}\:^{\mathrm{2}} \mathrm{3}{u}\:} \\ $$$$=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{3}{u}}{\mathrm{2sin}\:\mathrm{3}{u}}\right).\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{5tan}\:\mathrm{4}{u}\right)}{\mathrm{sin}\:\mathrm{3}{u}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{5tan}\:\mathrm{4}{u}\right)}{\mathrm{5tan}\:\mathrm{4}{u}}\:×\frac{\mathrm{5tan}\:\mathrm{4}{u}}{\mathrm{sin}\:\mathrm{3}{u}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\frac{\mathrm{20}}{\mathrm{3}}=\frac{\mathrm{10}}{\mathrm{3}} \\ $$

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