Question Number 114088 by bemath last updated on 17/Sep/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}{x}\:\mathrm{cot}\:\left(\frac{\mathrm{2}}{{x}}\right)−\mathrm{3cot}\:\left(\frac{\mathrm{2}}{{x}}\right)}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}} \\ $$ Answered by Olaf last updated on 17/Sep/20 $$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}}{{x}}\right)^{\mathrm{2}} }{\left(\frac{\mathrm{2}}{{x}}\right)}\left[\frac{\mathrm{2}{x}−\mathrm{3}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right]…
Question Number 114079 by bemath last updated on 17/Sep/20 Answered by john santu last updated on 17/Sep/20 $${setting}\:\frac{\mathrm{1}}{{x}}\:=\:{m}\:\wedge{m}\rightarrow\mathrm{0} \\ $$$$\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{m}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{4}{m}}{\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{m}\right)\mathrm{sin}\:{m}}\:= \\ $$$$\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{m}^{\mathrm{2}}…
Question Number 114075 by bemath last updated on 17/Sep/20 Answered by john santu last updated on 17/Sep/20 $${setting}\:\frac{\mathrm{1}}{{x}}={b}\:\wedge{b}\rightarrow\mathrm{0} \\ $$$$\underset{{b}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:{b}}\:−\:\frac{\mathrm{cos}\:{b}}{\mathrm{sin}\:{b}}\right)\:=\:\underset{{b}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{b}\right)}{\mathrm{sin}\:{b}}\right) \\ $$$$\:\:=\:\mathrm{0}…
Question Number 114028 by AbhishekBasnet last updated on 16/Sep/20 Commented by bemath last updated on 17/Sep/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{{x}^{\mathrm{2}} +{x}}\:−\sqrt{{x}}\:=\:\frac{\mathrm{1}−\mathrm{0}}{\mathrm{2}.\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by Olaf last…
Question Number 114020 by bemath last updated on 16/Sep/20 $$\underset{{d}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{6}{d}}\:\mathrm{cos}\:\left(\frac{\mathrm{2}}{\:\sqrt{{d}}}\right)\:\mathrm{sin}\:\left(\frac{\mathrm{5}}{\:\sqrt{{d}}}\right)\:?\: \\ $$$$ \\ $$ Answered by Olaf last updated on 17/Sep/20 $$\sqrt{\mathrm{6}{d}}\mathrm{cos}\left(\frac{\mathrm{2}}{\:\sqrt{{d}}}\right)\mathrm{sin}\left(\frac{\mathrm{5}}{\:\sqrt{{d}}}\right)\:\underset{\infty} {\sim}\:\sqrt{\mathrm{6}{d}}\left(\mathrm{1}−\frac{\mathrm{2}}{{d}}\right)\left(\frac{\mathrm{5}}{\:\sqrt{{d}}}\right) \\…
Question Number 179503 by cortano1 last updated on 30/Oct/22 $$\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)−\frac{\pi}{\mathrm{4}}}\:−\frac{\mathrm{2}}{\mathrm{x}−\mathrm{1}}\right)\:=? \\ $$ Commented by CElcedricjunior last updated on 30/Oct/22 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\boldsymbol{\mathrm{x}}−\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{arctan}}\left(\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\pi}/\mathrm{2}}{\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{arctanx}}−\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)}\right)=\frac{\mathrm{0}}{\mathrm{0}}=\boldsymbol{\mathrm{FI}} \\ $$$$\boldsymbol{{to}}\:\boldsymbol{{apply}}\:\boldsymbol{{hospital}}…
Question Number 179390 by greougoury555 last updated on 29/Oct/22 $$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{5}}{\:\sqrt{{x}^{\mathrm{3}} −\mathrm{1}}}\right)^{\sqrt{\frac{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left(\mathrm{4}{x}−\mathrm{1}\right)}{\mathrm{4}}}} =? \\ $$ Answered by cortano1 last updated on 29/Oct/22 Terms of…
Question Number 179366 by mathlove last updated on 28/Oct/22 $$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{{sin}\frac{{x}}{\mathrm{2}}−\mathrm{1}}{{x}−\pi}=? \\ $$ Commented by mahdipoor last updated on 28/Oct/22 $$={D}_{{x}} \left({sin}\frac{{x}}{\mathrm{2}}\right)_{{x}=\pi} =\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\frac{{x}}{\mathrm{2}}\right)_{{x}=\pi} =\mathrm{0} \\…
Question Number 113786 by bemath last updated on 15/Sep/20 Answered by bobhans last updated on 15/Sep/20 $${recall}\:\mathrm{tan}\:\mathrm{6}{x}\:=\:\mathrm{6}{x}+\frac{\left(\mathrm{6}{x}\right)^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}\left(\mathrm{6}{x}\right)^{\mathrm{5}} }{\mathrm{15}}+… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\left(\frac{\mathrm{4}}{{x}}\right)=\mathrm{1}−\frac{\left(\frac{\mathrm{4}}{{x}}\right)^{\mathrm{2}} }{\mathrm{2}!}+\frac{\left(\frac{\mathrm{4}}{{x}}\right)^{\mathrm{4}} }{\mathrm{4}!}−… \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 113742 by bemath last updated on 15/Sep/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} −{e}−\frac{{ex}}{\mathrm{2}}}{{x}^{\mathrm{2}} }\:=?\: \\ $$ Answered by bobhans last updated on 15/Sep/20 $${L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} −{e}−\frac{{ex}}{\mathrm{2}}}{{x}^{\mathrm{2}}…