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}}…
Question Number 179271 by cortano1 last updated on 27/Oct/22 $$\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\sqrt[{\mathrm{3}}]{\mathrm{n}^{\mathrm{3}} +\frac{\mathrm{n}}{\mathrm{3}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{n}}\right)\:\right]^{\mathrm{n}^{\mathrm{3}} } =? \\ $$$$\:\: \\ $$ Answered by mr W last updated on…
Question Number 113651 by bemath last updated on 14/Sep/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}+{e}^{−{x}} }\right).\frac{\mathrm{1}}{\mathrm{3}{x}}\:=\:? \\ $$ Answered by john santu last updated on 14/Sep/20 $${by}\:{Taylor}\:{series}\: \\ $$$${let}\:{f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+{e}^{−{x}}…
Question Number 113601 by bobhans last updated on 14/Sep/20 Answered by bemath last updated on 14/Sep/20 Commented by bobhans last updated on 14/Sep/20 $$\mathrm{santuyy} \\…
Question Number 113598 by bobhans last updated on 14/Sep/20 Answered by bemath last updated on 14/Sep/20 $${let}\:{me}\:{solve} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{3}\:}]{\mathrm{27}{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}} }\right)}−\sqrt[{\mathrm{4}\:}]{\mathrm{64}{x}^{\mathrm{4}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{3}} }\right)}\:= \\…
Question Number 113576 by bemath last updated on 14/Sep/20 $$\:\:\:\underset{{r}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{r}^{\mathrm{2}} +\mathrm{2}{r}}\:−\sqrt[{\mathrm{3}\:}]{\mathrm{8}{r}^{\mathrm{3}} +\mathrm{4}{r}^{\mathrm{2}} }\:=? \\ $$ Answered by bemath last updated on 14/Sep/20 $$\Leftrightarrow\:\underset{{r}\rightarrow\infty} {\mathrm{lim}}{r}\left(\sqrt{\mathrm{4}+\frac{\mathrm{2}}{{r}}}−\sqrt[{\mathrm{3}\:}]{\mathrm{8}+\frac{\mathrm{4}}{{r}}}\right)\:=…