Question Number 172715 by mathlove last updated on 30/Jun/22 $$\Omega=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}\left({x}!\right)}{\:\sqrt[{{x}}]{\mathrm{1}+{x}}−{e}}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 107163 by bobhans last updated on 09/Aug/20 Answered by john santu last updated on 09/Aug/20 $$\:\:\:\trianglerighteq\mathrm{JS}\trianglelefteq \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{cos}\:\mathrm{x}\:.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}\sqrt{\mathrm{1}+\frac{\mathrm{5}}{\mathrm{x}}}}{\mathrm{2}\left(\sqrt[{\mathrm{5}}]{\mathrm{1}−\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{8}}}−\mathrm{1}\right)}= \\ $$$$\mathrm{1}\:×\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}\sqrt{\mathrm{1}+\frac{\mathrm{5}}{\mathrm{x}}}}{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{40}}−\mathrm{1}\right)}=…
Question Number 107030 by bemath last updated on 08/Aug/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{2}+\frac{\mathrm{3}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{4}{x}−\mathrm{2}}} \\ $$ Commented by kaivan.ahmadi last updated on 08/Aug/20 $${y}=\left(\mathrm{2}+\frac{\mathrm{3}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{4}{x}−\mathrm{2}}} \Rightarrow{lim}_{{x}\rightarrow\infty} {lny}={lim}_{{x}\rightarrow\infty} \frac{{ln}\left(\mathrm{2}+\frac{\mathrm{3}}{{x}}\right)}{\mathrm{4}{x}−\mathrm{2}}\sim \\…
Question Number 107028 by bemath last updated on 08/Aug/20 $$\:\:\:@{bemath}@ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{2}+\mathrm{3}{x}\right)^{\frac{\mathrm{1}}{\mathrm{4}{x}−\mathrm{2}}} \\ $$ Commented by kaivan.ahmadi last updated on 08/Aug/20 $${y}=\left(\mathrm{2}+\mathrm{3}{x}\right)^{\frac{\mathrm{1}}{\mathrm{4}{x}−\mathrm{2}}} \Rightarrow{lny}=\frac{{ln}\left(\mathrm{2}+\mathrm{3}{x}\right)}{\mathrm{4}{x}−\mathrm{2}}\Rightarrow{lim}_{{x}\rightarrow\infty} {lny}=…
Question Number 41454 by Fawomath last updated on 07/Aug/18 $$\mathrm{Evaluate}\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {k}^{\mathrm{3}} \\ $$ Commented by maxmathsup by imad last updated on 07/Aug/18 $${let}\:{S}\left({x}\right)=\sum_{{k}=\mathrm{0}}…
Question Number 41431 by Tawa1 last updated on 07/Aug/18 Commented by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18 $${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} +\mathrm{4}{lnx}+{ln}\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{3}} \\ $$$$\left.\mathrm{1}\right){D}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{3} \\ $$$${finding}\:{zeroes}\:{for}\:{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{3}…
Question Number 106956 by bemath last updated on 08/Aug/20 Answered by 1549442205PVT last updated on 08/Aug/20 $$\frac{\mathrm{cos2x}\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)+\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{cos2x}\left(\mathrm{2}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)+\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }…
Question Number 172468 by mathlove last updated on 27/Jun/22 Commented by mokys last updated on 28/Jun/22 $$=\:\frac{\left(\mathrm{3}!!\right)^{\mathrm{3}!!} \:−\:\left(\mathrm{3}^{\mathrm{3}!!} \right)}{\mathrm{3}^{\mathrm{3}!} \:−\:\mathrm{3}^{\mathrm{6}!!} }\:=\:\frac{\mathrm{3}^{\mathrm{3}} \:−\:\mathrm{3}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}!} \:−\:\mathrm{3}^{\mathrm{6}!!} }\:=\:\mathrm{0}…
Question Number 106883 by john santu last updated on 07/Aug/20 $$\:\:\:\:\:\:\:\lozenge\mathrm{JS}\blacklozenge \\ $$$$\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}+\mathrm{1}−\sqrt{−\mathrm{x}}}{\:\sqrt{\mathrm{x}+\mathrm{1}}}\:?\: \\ $$ Answered by bemath last updated on 07/Aug/20 $$\:\:\:\:\:\:\:@\mathrm{bemath}@…
Question Number 172409 by cortano1 last updated on 26/Jun/22 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{sin}\:{x}}]{\mathrm{1}+\mathrm{2sin}\:{x}}−\sqrt[{\mathrm{tan}\:{x}}]{\mathrm{1}+\mathrm{tan}\:{x}}}{\mathrm{2}{x}^{\mathrm{3}} }=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com