Question Number 182256 by mathlove last updated on 06/Dec/22 Answered by Ar Brandon last updated on 06/Dec/22 $$\mathscr{L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{m}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\underset{{k}=\mathrm{1}} {\overset{{mr}} {\sum}}\frac{{mn}^{\mathrm{2}} }{\left({m}^{\mathrm{2}}…
Question Number 51095 by Tinkutara last updated on 23/Dec/18 Commented by prakash jain last updated on 25/Dec/18 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left[{x}\right]^{\mathrm{2}} −\left[{x}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} }…
Question Number 116588 by bemath last updated on 05/Oct/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4x}}{\mathrm{2}\:\mathrm{cosec}\:\mathrm{x}\:\left(\mathrm{1}−\sqrt{\mathrm{cos}\:\mathrm{x}}\right)}\:=? \\ $$ Answered by bobhans last updated on 05/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4x}.\mathrm{sin}\:\mathrm{x}}{\mathrm{2}\left(\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 116576 by bemath last updated on 05/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} \:=? \\ $$ Commented by mathmax by abdo last updated on 05/Oct/20 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}}…
Question Number 51005 by Tawa1 last updated on 23/Dec/18 Commented by maxmathsup by imad last updated on 23/Dec/18 $${let}\:{A}\left(\theta\right)\:=\frac{{sin}\left(\theta−\frac{\pi}{\mathrm{6}}\right)}{\:\sqrt{\mathrm{3}}\:−\mathrm{2}{cos}\theta}\:\:{changement}\:\:\theta\:−\frac{\pi}{\mathrm{6}}\:={x}\:{give} \\ $$$${A}\left(\theta\right)\:=\frac{{sinx}}{\:\sqrt{\mathrm{3}}−\mathrm{2}{cos}\left({x}+\frac{\pi}{\mathrm{6}}\right)}\:=\frac{{sinx}}{\:\sqrt{\mathrm{3}}−\mathrm{2}\left({cosx}\:{cos}\left(\frac{\pi}{\mathrm{6}}\right)−{sinx}\:{sin}\left(\frac{\pi}{\mathrm{6}}\right)\right)} \\ $$$$=\:\frac{{sinx}}{\:\sqrt{\mathrm{3}}−\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cos}−\frac{\mathrm{1}}{\mathrm{2}}{sinx}\right)}\:=\frac{{sinx}}{\:\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{cosx}\:+{sinx}}\:{but}\:{lim}_{\theta\rightarrow\frac{\pi}{\mathrm{6}}} \:\:={lim}_{{x}\rightarrow\mathrm{0}} \frac{{sinx}}{\:\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{cosx}+{sinx}}…
Question Number 182077 by cortano1 last updated on 04/Dec/22 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}\:\mathrm{ln}\:\left(\frac{\sqrt[{\mathrm{4}}]{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{2}}}{\:\sqrt[{\mathrm{4}}]{\mathrm{16x}^{\mathrm{2}} +\mathrm{2x}}\:−\sqrt{\mathrm{x}}}\:\right)=? \\ $$ Answered by SEKRET last updated on 04/Dec/22 $$\:\:\frac{\mathrm{7}}{\mathrm{16}} \\ $$…
Question Number 50954 by afachri last updated on 22/Dec/18 $$\mathrm{i}\:\mathrm{was}\:\mathrm{evaluating}\: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}\left(\sqrt[{{x}}]{{x}^{} }\:−\:\mathrm{1}\right)}{\mathrm{log}\:{x}} \\ $$$$\mathrm{and}\:\mathrm{got}\:\mathrm{0}\:\mathrm{as}\:\mathrm{the}\:\mathrm{product}.\:\mathrm{is}\:\mathrm{it}\:\mathrm{true},\:\mathrm{My}\:\mathrm{Fellows}\:?? \\ $$ Commented by Abdo msup. last updated on…
Question Number 116451 by bemath last updated on 04/Oct/20 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{n}}]{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)…\left(\mathrm{n}+\mathrm{n}\right)}}{\mathrm{n}}\:=? \\ $$ Commented by Olaf last updated on 04/Oct/20 $$\boldsymbol{\mathrm{Sorry}}\:\boldsymbol{\mathrm{I}}'\boldsymbol{\mathrm{m}}\:\boldsymbol{\mathrm{wrong}}\:! \\ $$$$\mathrm{I}\:\boldsymbol{\mathrm{calculated}}\:\boldsymbol{\mathrm{another}}\:\boldsymbol{\mathrm{expression}}\:! \\ $$$$!??!\:\boldsymbol{\mathrm{I}}'\boldsymbol{\mathrm{m}}\:\boldsymbol{\mathrm{tired}}\:!…
Question Number 116417 by bemath last updated on 03/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\mathrm{x}\:\mathrm{sin}\:\mathrm{x}}\:−\sqrt{\mathrm{cos}\:\mathrm{2x}}\:}{\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\:=? \\ $$ Answered by Bird last updated on 03/Oct/20 $${f}\left({x}\right)=\frac{\sqrt{\mathrm{1}+{xsinx}}−\sqrt{{cos}\left(\mathrm{2}{x}\right)}}{{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$${we}\:{hsve}\:\sqrt{\mathrm{1}+{xsinx}}\sim\sqrt{\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 181901 by SATYA121993 last updated on 01/Dec/22 Terms of Service Privacy Policy Contact: info@tinkutara.com