Question Number 59961 by bhanukumarb2@gmail.com last updated on 16/May/19 Commented by bhanukumarb2@gmail.com last updated on 16/May/19 $${i}\:{know}\:{one}\:{mthd}\:{by}\:{stirling}\:{approximation} \\ $$$${plz}\:{try}\:{any}\:{other}\:{mthd} \\ $$$${doubt}\:{we}\:{can}\:{use}\:{cesaro}\:{theoram}\:{or}\:{nt}???? \\ $$$$ \\ $$…
Question Number 190996 by cortano12 last updated on 16/Apr/23 $$\:\:\:\mathrm{If}\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{3x}}−\mathrm{3}}{\:\sqrt{\mathrm{2x}−\mathrm{4}}−\sqrt{\mathrm{2}}}\:=\:\mathrm{A} \\ $$$$\:\:\:\mathrm{and}\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{3x}}+\mathrm{x}−\mathrm{6}}{\:\sqrt{\mathrm{4x}−\mathrm{8}}+\mathrm{1}−\mathrm{x}}\:=\:\mathrm{pA} \\ $$$$\:\:\:\mathrm{then}\:\mathrm{p}\:=?\: \\ $$ Commented by mustafazaheen last updated on 16/Apr/23…
Question Number 190998 by mathlove last updated on 16/Apr/23 Commented by mr W last updated on 16/Apr/23 $${is}\:{it}\:{normal}\:{when}\:{somebody}\:{asks} \\ $$$${you}\:{directly}\:{a}\:{question}\:{but}\:{you}\:{totally} \\ $$$${irgnore}\:{it}\:{without}\:{giving}\:{any}\:{reply}? \\ $$$${i}\:{find}\:{this}\:{behaviour}\:{at}\:{least}\:{impolite}. \\…
Question Number 125394 by Mammadli last updated on 10/Dec/20 $$\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{lim}}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{3}} }+…+\frac{\mathrm{2}\boldsymbol{{n}}−\mathrm{1}}{\mathrm{2}^{\boldsymbol{{n}}} }\right)=? \\ $$ Answered by Dwaipayan Shikari last updated on 10/Dec/20 $$\underset{{n}=\mathrm{1}}…
Question Number 59835 by bhanukumarb2@gmail.com last updated on 15/May/19 Answered by tanmay last updated on 15/May/19 $${T}_{{r}} =\frac{\left({cos}\mathrm{2}\theta\right)^{\mathrm{2}{r}−\mathrm{1}} }{\mathrm{2}{r}−\mathrm{1}} \\ $$$${l}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\Sigma\frac{\left({cos}\mathrm{2}\theta\right)^{\mathrm{2}{r}−\mathrm{1}} }{\mathrm{2}{r}−\mathrm{1}} \\ $$$${l}=\frac{\left({cos}\mathrm{2}\theta\right)^{\mathrm{1}}…
Question Number 190824 by cortano12 last updated on 12/Apr/23 $$\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{2}\sqrt{\mathrm{tan}\:\mathrm{x}}}{\mathrm{2cos}\:\mathrm{x}−\mathrm{2sin}\:\mathrm{x}}\:=? \\ $$ Commented by 0670322918 last updated on 12/Apr/23 $$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{{tan}^{\mathrm{2}} \left({x}\right)+\mathrm{1}−\mathrm{2}+\mathrm{2}−\mathrm{2}\sqrt{{tan}\left({x}\right)}}{{cos}\left({x}\right)\left[\mathrm{1}−{tan}\left({x}\right)\right]}= \\…
Question Number 125292 by Kurbanklichevs last updated on 09/Dec/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)+{n}}{{x}^{\mathrm{2}} }\:\:\:\:\:{n}\in\mathbb{N} \\ $$$${Calculate}.\:\left({Without}\:{a}\:{L}'{Hopital}'{s}\:{rule}\right) \\ $$ Answered by Dwaipayan Shikari last updated on 09/Dec/20…
Question Number 125282 by Mammadli last updated on 09/Dec/20 $$\mathrm{1}.\:\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\sqrt[{\boldsymbol{{n}}}]{\frac{\mathrm{2}\boldsymbol{{n}}−\mathrm{1}}{\mathrm{2}\boldsymbol{{n}}+\mathrm{1}}}=? \\ $$$$\mathrm{2}.\:\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{lim}}}\sqrt[{\boldsymbol{{n}}}]{\frac{\mathrm{2}\boldsymbol{{n}}−\mathrm{1}}{\mathrm{2}\boldsymbol{{n}}+\mathrm{1}}}=? \\ $$ Answered by mathmax by abdo last updated on…
Question Number 190788 by mnjuly1970 last updated on 11/Apr/23 $$ \\ $$$$\:\:\:\:\mathrm{calculate}\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\Omega=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\:\mathrm{1}}{\left({n}−{k}\right)!.\left({n}+{k}\:\right)!} \\ $$$$ \\ $$ Answered by aleks041103…
Question Number 59694 by meme last updated on 13/May/19 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{cos}\left(\sqrt{\left.\mid{x}\mid\right)−\mathrm{1}}\right.}{{x}}=? \\ $$ Commented by maxmathsup by imad last updated on 13/May/19 $${let}\:{A}\left({x}\right)\:=\frac{{cos}\left(\sqrt{\mid{x}\mid}\right)−\mathrm{1}}{{x}}\:\:\:\:{we}\:{have}\:{cos}\left({u}\right)\:\sim\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\left(\:\:{u}\in{V}\left(\mathrm{0}\right)\right)\:\Rightarrow \\…