Question Number 209926 by depressiveshrek last updated on 26/Jul/24 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}+…+\frac{\mathrm{1}}{\mathrm{4}{n}} \\ $$ Commented by Frix last updated on 26/Jul/24 $$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}{n}+{k}}\:={H}_{\mathrm{4}{n}} −{H}_{\mathrm{3}{n}} \\…
Question Number 209810 by mnjuly1970 last updated on 22/Jul/24 $$ \\ $$$$\:\:\:\mathrm{lim}_{\:\mathrm{x}\rightarrow\mathrm{0}} \:\frac{\:\left(\mathrm{1}+\:\mathrm{x}\:\right)^{\frac{\mathrm{1}}{\mathrm{x}}} −\mathrm{e}}{\mathrm{x}}\:=\:? \\ $$$$ \\ $$ Answered by mr W last updated on…
Question Number 209434 by justenspi last updated on 10/Jul/24 $${Help} \\ $$ Commented by justenspi last updated on 10/Jul/24 Commented by Berbere last updated on…
Question Number 209356 by mnjuly1970 last updated on 08/Jul/24 $$ \\ $$$$\:\:\:\:\:{Evaluate}\:: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\mathrm{lim}_{\:{n}\rightarrow\infty} \:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\:{cos}\:\left(\frac{\mathrm{2}^{\:{k}} .\pi}{\mathrm{2}^{\:{n}} \:−\mathrm{1}}\:\right)\:\:=\:?\:\:\:\:\:\:\:\:\:\: \\ $$$$…
Question Number 209220 by alcohol last updated on 04/Jul/24 Answered by Berbere last updated on 04/Jul/24 $${SAB}\:,{SAC}\:\:\&{ABC}\:{c}\:{est}\:{claire}\:\left({AS}\right)\bot\left({ABC}\right)\:\&{ABC}\:{rectangle} \\ $$$$\Rightarrow{SA}^{\mathrm{2}} +{AB}^{\mathrm{2}} ={SB}^{\mathrm{2}} ;{SA}^{\mathrm{2}} +{AC}^{\mathrm{2}} ={SC}^{\mathrm{2}} ;{CA}^{\mathrm{2}}…
Question Number 209116 by alcohol last updated on 02/Jul/24 Answered by A5T last updated on 02/Jul/24 $${WLOG},\:{let}\:{a}\geqslant{b}\geqslant{c} \\ $$$$\mathrm{1}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\leqslant\frac{\mathrm{3}}{{c}}\Rightarrow{c}\leqslant\mathrm{3} \\ $$$${when}\:{c}=\mathrm{3};\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\leqslant\frac{\mathrm{2}}{{b}}\Rightarrow{b}\leqslant\mathrm{3} \\ $$$${b}=\mathrm{3}\Rightarrow{a}=\mathrm{3};\:{b}=\mathrm{2}\Rightarrow{a}=\mathrm{6}\:\:\:\rightarrow\leftarrow \\ $$$$\Rightarrow\left({a},{b},{c}\right)=\left(\mathrm{3},\mathrm{3},\mathrm{3}\right)…
Question Number 208816 by alcohol last updated on 23/Jun/24 Commented by alcohol last updated on 23/Jun/24 $${please}\:{help} \\ $$ Terms of Service Privacy Policy Contact:…
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Question Number 208093 by depressiveshrek last updated on 04/Jun/24 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{{n}}\left(\left({a}+\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}} +\left({a}+\frac{\mathrm{2}}{{n}}\right)^{\mathrm{2}} +…+\left({a}+\frac{{n}−\mathrm{1}}{{n}}\right)^{\mathrm{2}} \right) \\ $$ Answered by MM42 last updated on 04/Jun/24 $$={lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\underset{{i}=\mathrm{1}}…
Question Number 207816 by efronzo1 last updated on 27/May/24 $$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}−\sqrt{\mathrm{x}^{\mathrm{3}} −\mathrm{4}}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{4}}−\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{2}}} \\ $$ Commented by Frix last updated on 27/May/24 $$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{0} \\…