Question Number 114637 by bobhans last updated on 20/Sep/20 $$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\frac{\pi}{\mathrm{2}}−\mathrm{cos}\:^{−\mathrm{1}} \left(\mathrm{2}{x}−\pi\right)}{\mathrm{1}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\pi}\right)}\:? \\ $$ Answered by bemath last updated on 20/Sep/20 $${setting}\:{x}=\frac{\pi}{\mathrm{2}}+{p}\rightarrow\mathrm{2}{x}=\pi+\mathrm{2}{p} \\ $$$$\underset{{p}\rightarrow\mathrm{0}}…
Question Number 114543 by bemath last updated on 19/Sep/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \:\mathrm{arctan}\:\left(\mathrm{2}{x}\right)}{{x}\:\mathrm{cos}\:{x}+\mathrm{tan}\:{x}}\:? \\ $$ Answered by bobhans last updated on 19/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{{x}\:\mathrm{cos}\:{x}+\mathrm{tan}\:{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 114533 by 675480065 last updated on 19/Sep/20 Answered by abdomsup last updated on 19/Sep/20 $${u}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right) \\ $$$$\left.{a}\right){let}\:\varphi\left({x}\right)={x}−{ln}\left(\mathrm{1}+{x}\right) \\ $$$${with}\:{x}\geqslant\mathrm{0}\:{we}\:{have}\:\varphi\left(\mathrm{0}\right)=\mathrm{0}…
Question Number 179950 by mathlove last updated on 04/Nov/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)+\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)^{\frac{\mathrm{2}}{\mathrm{2}}} +\left(\mathrm{1}+\frac{\mathrm{3}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\centerdot\centerdot\centerdot\centerdot+\left(\mathrm{1}+\frac{{x}}{{x}}\right)^{\frac{\mathrm{1}}{{x}}} \right]=? \\ $$ Commented by mahdipoor last updated on 04/Nov/22 $$ \\…
Question Number 114411 by bemath last updated on 19/Sep/20 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} −\left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{2}{x}}\:=? \\ $$ Answered by john santu last updated on 19/Sep/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}}…
Question Number 114405 by bemath last updated on 19/Sep/20 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}−\mathrm{sin}\:^{\mathrm{2}} {x}−\sqrt{\mathrm{3}}}\:? \\ $$ Answered by bobhans last updated on 19/Sep/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{3}}\left(\mathrm{cos}\:{x}−\mathrm{1}\right)−\mathrm{sin}\:^{\mathrm{2}} {x}}\:= \\…
Question Number 114299 by bemath last updated on 18/Sep/20 $$\:{solve}\:\underset{{p}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{p}^{\mathrm{2}} +\mathrm{2}}{{p}+\mathrm{2}}\right)^{\frac{{p}+\mathrm{2}}{{p}^{\mathrm{2}} +\mathrm{2}}} \:? \\ $$ Commented by mohammad17 last updated on 18/Sep/20 $${put}:{k}=\frac{{p}+\mathrm{2}}{{p}^{\mathrm{2}} +\mathrm{2}}\Rightarrow{k}\rightarrow\mathrm{0}…
Question Number 114259 by bobhans last updated on 18/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:\mathrm{arc}\:\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}{{x}\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}\:? \\ $$ Answered by bemath last updated on 18/Sep/20 $${by}\:{L}'{Hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{arc}\:\mathrm{sin}\:\left({x}^{\mathrm{2}}…
Question Number 114239 by bemath last updated on 18/Sep/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\left(\frac{\mathrm{4}}{{x}}\right)}−\mathrm{1}}{\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)−\mathrm{cos}\:\left(\frac{\mathrm{4}}{{x}}\right)}\:? \\ $$ Commented by bemath last updated on 18/Sep/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{1}−\frac{\mathrm{8}}{{x}^{\mathrm{2}} }}−\mathrm{1}}{\left(\mathrm{1}−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right)−\left(\mathrm{1}−\frac{\mathrm{8}}{{x}^{\mathrm{2}} }\right)}\:=…
Question Number 179657 by mathlove last updated on 31/Oct/22 $${prove}\:{that} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}}×\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}}×\centerdot\centerdot\centerdot\centerdot\centerdot\underset{{n}\:{term}} {\underbrace{\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}+}…..+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}}}}}\right)=\frac{\mathrm{2}}{\pi} \\ $$ Commented by mr W last updated on 01/Nov/22 $${the}\:{question}\:{should}\:{be}:…