Question Number 117164 by bobhans last updated on 10/Oct/20 $$\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{n}+\mathrm{9}}{\mathrm{2n}−\mathrm{1}}\right)^{\mathrm{n}} =? \\ $$ Answered by bemath last updated on 10/Oct/20 $$\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{2n}−\mathrm{1}+\mathrm{10}−\mathrm{n}}{\mathrm{2n}−\mathrm{1}}\right)^{\mathrm{n}} = \\…
Question Number 51617 by Saorey last updated on 29/Dec/18 $$\mathrm{A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} −{e}^{−{x}} −\mathrm{2}{x}}{{x}^{\mathrm{3}} }\left({without}\:\right. \\ $$$$\left.{using}\:{L}'\mathrm{hopital}\:\mathrm{rule}\right) \\ $$$$ \\ $$ Commented by aseerimad last updated…
Question Number 117147 by bemath last updated on 10/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{cosh}\:\left(\mathrm{2x}\right)}{\mathrm{cosh}\:\left(\mathrm{x}\right)}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} \:=? \\ $$ Answered by Olaf last updated on 10/Oct/20 $$\mathrm{cosh}{u}\:\underset{\mathrm{0}} {\sim}\:\mathrm{1}+\frac{{u}^{\mathrm{2}} }{\mathrm{2}}…
Question Number 182659 by mathlove last updated on 12/Dec/22 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{{n}}]{{n}!}}{{n}}=? \\ $$ Answered by Frix last updated on 12/Dec/22 $${n}!\sim\frac{{n}^{{n}} \sqrt{\mathrm{2}\pi{n}}}{\mathrm{e}^{{n}} } \\ $$$$\underset{{n}\rightarrow\infty}…
Question Number 182649 by amin96 last updated on 12/Dec/22 $$\boldsymbol{{f}}\left(\boldsymbol{{x}};\boldsymbol{{y}}\right)=\frac{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} }{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{4}} }\:\:\:\:\: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{f}}\left(\boldsymbol{{x}};\boldsymbol{{y}}\right)\right)=? \\ $$$$\underset{\boldsymbol{{y}}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{x}\rightarrow\infty} {\mathrm{lim}}\boldsymbol{{f}}\left(\boldsymbol{{x}};\boldsymbol{{y}}\right)\right)=? \\ $$ Answered…
Question Number 182646 by mathlove last updated on 12/Dec/22 $${prove}\:{that} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\sqrt[{{n}+\mathrm{1}}]{\left({n}+\mathrm{1}\right)!\centerdot\left(\mathrm{2}{n}+\mathrm{1}\right)!!}}{{n}+\mathrm{1}}−\frac{\sqrt[{{n}}]{{n}!\centerdot\left(\mathrm{2}{n}−\mathrm{1}\right)!!}}{{n}}\right]=\frac{\mathrm{2}}{{e}^{\mathrm{2}} } \\ $$ Commented by mathlove last updated on 13/Dec/22 $$??? \\…
Question Number 182629 by universe last updated on 12/Dec/22 $$\:\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{2}} {\overset{\infty} {\prod}}\:\boldsymbol{\mathrm{e}}\left(\mathrm{1}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }\right)^{\boldsymbol{\mathrm{n}}^{\mathrm{2}} } \:=\:\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 117078 by bemath last updated on 09/Oct/20 $$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2n}}\right).\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{2n}}\right).\mathrm{tan}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2n}}\right)…\mathrm{tan}\:\left(\frac{\mathrm{n}\pi}{\mathrm{2n}}\right)\right]^{\frac{\mathrm{1}}{\mathrm{n}}} =?\: \\ $$ Answered by Dwaipayan Shikari last updated on 09/Oct/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{r}=\mathrm{1}} {\overset{{n}}…
Question Number 117052 by bemath last updated on 09/Oct/20 $$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4x}+\sqrt{\mathrm{4x}+\sqrt{\mathrm{4x}+\sqrt{\mathrm{4x}}}}}\:−\:\sqrt{\mathrm{4x}}\:=? \\ $$ Answered by bobhans last updated on 09/Oct/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4x}\left(\mathrm{1}+\frac{\sqrt{\mathrm{4x}+\sqrt{\mathrm{4x}+\sqrt{\mathrm{4x}}}}}{\mathrm{4x}}\right)}\:−\sqrt{\mathrm{4x}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4x}}\:\left[\:\sqrt{\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{4x}}+\sqrt{\frac{\mathrm{1}}{\left(\mathrm{4x}\right)^{\mathrm{3}}…
Question Number 117037 by bemath last updated on 09/Oct/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}\:−\:\sqrt{\mathrm{x}}\:=?\: \\ $$ Answered by Lordose last updated on 09/Oct/20 $$\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by…