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)} \\ $$$${we}\:{hsve}\:\sqrt{\mathrm{1}+{xsinx}}\sim\sqrt{\mathrm{1}+{x}^{\mathrm{2}}…
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Question Number 116331 by bobhans last updated on 03/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3x}\left(\mathrm{cos}\:\mathrm{7x}−\mathrm{cos}\:\mathrm{3x}\right)}{\:\sqrt{\mathrm{tan}\:\mathrm{2x}+\mathrm{1}}−\sqrt{\mathrm{sin}\:\mathrm{2x}+\mathrm{1}}}\:? \\ $$ Answered by bemath last updated on 03/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\sqrt{\mathrm{tan}\:\mathrm{2x}+\mathrm{1}}+\sqrt{\mathrm{sin}\:\mathrm{2x}+\mathrm{1}}\right)×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3x}\left(−\mathrm{2sin}\:\mathrm{5x}.\mathrm{sin}\:\mathrm{2x}\right)}{\mathrm{tan}\:\mathrm{2x}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2x}\right)} \\ $$$$=\:\mathrm{2}\:×\underset{{x}\rightarrow\mathrm{0}}…
Question Number 116317 by bemath last updated on 03/Oct/20 $$\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\sqrt{\mathrm{n}+\mathrm{1}}+\sqrt{\mathrm{n}+\mathrm{2}}+\sqrt{\mathrm{n}+\mathrm{3}}+…+\sqrt{\mathrm{2n}−\mathrm{1}}}{\mathrm{n}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\right)\:= \\ $$ Answered by Bird last updated on 03/Oct/20 $${U}_{{n}} \:=\frac{\mathrm{1}}{{n}^{\frac{\mathrm{3}}{\mathrm{2}}} }\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}}…
Question Number 116252 by bemath last updated on 02/Oct/20 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }\:=\:? \\ $$ Answered by MJS_new last updated on 02/Oct/20 $$\frac{\mathrm{e}^{{x}}…
Question Number 116167 by Ar Brandon last updated on 01/Oct/20 $$\mathrm{Show}\:\mathrm{that}\:\forall{n}\geqslant\mathrm{2}\:\mathrm{the}\:\mathrm{equation}\:{x}^{{n}} ={x}+{n} \\ $$$$\mathrm{admits}\:\mathrm{a}\:\mathrm{unique}\:\mathrm{solution}\:\mathrm{u}_{\mathrm{n}} \in\left(\mathrm{1},\mathrm{2}\right] \\ $$ Answered by 1549442205PVT last updated on 01/Oct/20 $$\mathrm{Put}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{n}}…