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 $${the}\:{question}\:{should}\:{be}:…
Question Number 48581 by Kiran bendkoli last updated on 25/Nov/18 $${f}\left({x}\right)=\left[\mathrm{tan}\left(\:\pi/\mathrm{4}+{x}\right)^{\mathrm{1}/{x}} \right] \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{k} \\ $$$${is}\:{con}\mathrm{ntinous}\:\mathrm{at}\:\mathrm{x}=\mathrm{0}\:\mathrm{then}\:\mathrm{k}=? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18…
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) \\…