Question Number 26751 by abdo imad last updated on 28/Dec/17 $${by}\:{using}\:{fourier}\:{serie}\:{find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Commented by abdo imad last updated on 30/Dec/17 $${we}\:{know}\:{that}\:/{x}/=\:\frac{\pi}{\mathrm{2}}\:−\frac{\mathrm{4}}{\pi}\:\sum_{{p}=\mathrm{0}}…
Question Number 26749 by abdo imad last updated on 28/Dec/17 $${let}\:{give}\:\:{S}_{{n}} \:=\:\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\:\frac{\mathrm{1}}{{i}^{\mathrm{2}} {j}^{\mathrm{2}} }\:\:\:{find}\:{lim}_{{n}−>\propto} \:\:{S}_{{n}} \:\:. \\ $$ Commented by abdo imad last updated…
Question Number 157803 by cortano last updated on 28/Oct/21 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:\mathrm{tan}\:\left(\mathrm{tan}\:{x}\right)−\left(\mathrm{tan}\:{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{6}} }\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 26722 by prakash jain last updated on 28/Dec/17 $$\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\mathrm{up}\:\mathrm{to}\:\infty\right)=? \\ $$$$\underset{{x}\rightarrow−\mathrm{1}^{−} } {\mathrm{lim}}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\mathrm{up}\:\mathrm{to}\:\infty\right)=? \\ $$ Terms of…
Question Number 92256 by jagoll last updated on 05/May/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}\:? \\ $$ Commented by $@ty@m123 last updated on 05/May/20 $${See}\:{Q}.\:{No}.\:\mathrm{87105} \\ $$ Commented by…
Question Number 92239 by jagoll last updated on 05/May/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mid\mathrm{cos}\:\mathrm{7x}\mid}{\mathrm{1}−\mid\mathrm{tan}\:\mathrm{5x}\mid}\:=\: \\ $$ Commented by john santu last updated on 05/May/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\mathrm{7x}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \:\mathrm{5x}}\:×\:\frac{\mathrm{1}+\mid\mathrm{tan}\:\mathrm{5x}\mid}{\mathrm{1}+\mid\mathrm{cos}\:\mathrm{7x}\mid\:}…
Question Number 92214 by hovero clinton last updated on 05/May/20 Commented by mathmax by abdo last updated on 05/May/20 $${for}\:{x}\rightarrow\mathrm{0}^{+} \:\:{we}\:{see}\:\frac{\mathrm{1}}{{lnx}}<\mathrm{0}\:\:\:{so}\:{we}\:{can}\:{t}\:{use}\:{expo}\:…!\:\:{perhaps}\:{the} \\ $$$${Q}\:{is}\:{gind}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:\left(−\frac{\mathrm{1}}{{lnx}}\right)^{\frac{\mathrm{1}}{{ln}\left({e}^{{x}}…
Question Number 92167 by john santu last updated on 05/May/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}−\mathrm{x}^{\mathrm{2}} \mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\:? \\ $$ Commented by mathmax by abdo last updated on 06/May/20 $${we}\:{have}\:{ln}^{'}…
Question Number 157688 by tounghoungko last updated on 26/Oct/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{ln}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)}\:−\frac{\mathrm{1}}{\mathrm{ln}\:\left({x}+\mathrm{1}\right)}\:\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 26591 by prakash jain last updated on 27/Dec/17 $$\mathrm{Prove} \\ $$$$\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)\mathrm{ln}\:\left(\mathrm{1}+{x}\right)=\mathrm{0} \\ $$ Commented by abdo imad last updated on 27/Dec/17 $${let}\:{put}\:\mathrm{1}+{x}={u}\:\:\:\Rightarrow\:{lim}_{{x}−>−\mathrm{1}^{+}…