Question Number 163658 by mathlove last updated on 09/Jan/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{{n}}]{\mathrm{sin}\:\frac{\pi}{\mathrm{2}{n}}×\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}}×\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{2}{n}}….×\mathrm{sin}\:\frac{\left({n}−\mathrm{1}\right)\pi}{{n}}}=? \\ $$ Answered by Ar Brandon last updated on 09/Jan/22 $$\mathcal{A}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{{n}}]{\mathrm{sin}\:\frac{\pi}{\mathrm{2}{n}}×\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}}×\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{2}{n}}….×\mathrm{sin}\:\frac{\left({n}−\mathrm{1}\right)\pi}{{n}}} \\ $$$$\mathrm{ln}\mathcal{A}=\underset{{n}\rightarrow\infty}…
Question Number 98114 by malwaan last updated on 11/Jun/20 $$\mathrm{1}\:\:\:\underset{\boldsymbol{{x}}\rightarrow\infty} {\boldsymbol{{lim}}}\:^{\mathrm{3}} \sqrt{\mathrm{5}\boldsymbol{{x}}^{\mathrm{3}} }\:=\:? \\ $$$$\mathrm{2}\:\:\underset{\boldsymbol{{x}}\rightarrow\infty} {\boldsymbol{{lim}}}\:\left(\mathrm{1}\:+\:\frac{\boldsymbol{{n}}}{\boldsymbol{{x}}\:+\:\boldsymbol{\alpha}}\right)^{\boldsymbol{{x}}} ;\boldsymbol{\alpha}\:\boldsymbol{{is}}\:\boldsymbol{{constant}} \\ $$ Answered by mahdi last updated on…
Question Number 163650 by mathlove last updated on 09/Jan/22 $$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{{x}^{\pi^{{x}} } −\pi^{{x}^{\pi} } }{{x}−\pi}=? \\ $$ Commented by Zaynal last updated on 09/Jan/22 $$\mathrm{is}'\mathrm{t}\:\mathrm{not}\:\mathrm{value}\:\mathrm{of}\:\mathrm{limit}…
Question Number 163642 by cortano1 last updated on 09/Jan/22 $$\:\:\:\:\:\:\underset{{x}\rightarrow\frac{\mathrm{2}}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\lfloor\mathrm{3}{x}\rfloor−\mathrm{3}{x}}{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}}\:=? \\ $$ Answered by mehdiAz last updated on 09/Jan/22 $$\: \\ $$$$\:\:\:\:\:\:\underset{{x}\rightarrow\frac{\mathrm{2}}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\lfloor\mathrm{3}{x}\rfloor−\mathrm{3}{x}}{\mathrm{9}{x}^{\mathrm{2}}…
Question Number 163609 by qaz last updated on 08/Jan/22 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}n}\left[\mathrm{A}−\mathrm{n}\left(\mathrm{H}_{\mathrm{n}} −\mathrm{lnn}−\gamma\right)\right]=\mathrm{B} \\ $$$$\mathrm{Find}\:\frac{\mathrm{A}}{\mathrm{B}}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163608 by qaz last updated on 08/Jan/22 $$\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{n}}{\mathrm{n}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{n}}{\mathrm{n}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }+…+\frac{\mathrm{n}}{\mathrm{n}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} } \\ $$$$\mathrm{Prove}::\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{4}} \left\{\frac{\mathrm{1}}{\mathrm{24}}−\mathrm{n}\left[\mathrm{n}\left(\frac{\pi}{\mathrm{4}}−\mathrm{A}_{\mathrm{n}} \right)−\frac{\mathrm{1}}{\mathrm{4}}\right]\right\}}=\mathrm{2016} \\ $$ Terms…
Question Number 32517 by abdo imad last updated on 26/Mar/18 $${calculatelim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\frac{{x}^{{sinx}} \:\:−\left({sinx}\right)^{{x}} }{{x}}\:. \\ $$ Commented by abdo imad last updated on 27/Mar/18…
Question Number 163582 by mathlove last updated on 08/Jan/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{7}^{{x}+\mathrm{2}} +\mathrm{6}^{{x}} }{\mathrm{3}^{\mathrm{2}{x}} −\mathrm{5}^{{x}} }=? \\ $$ Answered by cortano1 last updated on 08/Jan/22 $$\:\underset{{x}\rightarrow\infty}…
Question Number 98004 by Ar Brandon last updated on 11/Jun/20 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{sin}\left(\mathrm{n}\right)+\mathrm{4}^{\mathrm{n}} ×\frac{\mathrm{3}}{\mathrm{n}^{\mathrm{2}} }×\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\mathrm{4}}\right] \\ $$ Commented by bobhans last updated on 11/Jun/20 $$\underset{\mathrm{n}\rightarrow\infty}…
Question Number 32470 by daffa123 last updated on 25/Mar/18 $$\underset{{x}\rightarrow\infty} {{lim}}\:{x}\left({sec}\left(\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)\:−\:\mathrm{1}\right)=… \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com