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}} \\ $$ Answered by…
Question Number 182560 by universe last updated on 11/Dec/22 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:−\:\mathrm{cot}^{\mathrm{2}} {x}\:=\:\:? \\ $$ Answered by cortano1 last updated on 11/Dec/22 $$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{x}}{\mathrm{x}^{\mathrm{3}} }\:.\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 116970 by bemath last updated on 08/Oct/20 Answered by MJS_new last updated on 08/Oct/20 $$\mathrm{the}\:\mathrm{limit}\:{x}\rightarrow\mathrm{1}\:\mathrm{is}\:\mathrm{just}\:\mathrm{the}\:\mathrm{value}?! \\ $$$$=\sqrt{\frac{\mathrm{7}}{\mathrm{3}}} \\ $$ Commented by bemath last…
Question Number 116930 by bobhans last updated on 08/Oct/20 $$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{5}^{\sqrt{\mathrm{x}}} \:\mathrm{find}\:\mathrm{f}\:'\left(\mathrm{x}\right)\:\mathrm{by}\:\mathrm{using}\:\mathrm{limit} \\ $$$$\left(\mathrm{first}\:\mathrm{principal}\:\mathrm{derivative}\right) \\ $$ Commented by bobhans last updated on 08/Oct/20 Answered by 1549442205PVT…
Question Number 182453 by universe last updated on 09/Dec/22 Answered by qaz last updated on 10/Dec/22 $$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cot}\:{x}−\mathrm{cot}\:\mathrm{2}{x} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\mathrm{tan}\:\frac{\Psi}{\mathrm{2}^{{k}} }=\mathrm{tan}\:\Psi+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}^{{k}}…
Question Number 182441 by mathlove last updated on 09/Dec/22 Answered by JDamian last updated on 09/Dec/22 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{{xk}} \right)−\mathrm{ln}\:{n}}{{x}}=\mathrm{9} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\underset{{k}=\mathrm{1}} {\overset{{n}}…