Question Number 99676 by bemath last updated on 22/Jun/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{sin}\:\left(\mathrm{sin}\:\mathrm{x}\right)−\mathrm{x}}{\mathrm{x}\left(\mathrm{cos}\:\left(\mathrm{sin}\:\mathrm{x}\right)−\mathrm{1}\right)}?? \\ $$ Answered by bobhans last updated on 23/Jun/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{x}−\frac{\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{6}}\:−\mathrm{x}}{\mathrm{x}\left(\mathrm{1}−\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{2}}\:−\mathrm{1}\right)}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}}…
Question Number 34092 by rahul 19 last updated on 30/Apr/18 $$\underset{{n}} {{l}}\underset{} {{i}}\underset{\infty} {{m}}\:\left(\frac{\left({n}!\right)}{\left({nm}\right)^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} \\ $$ Commented by MJS last updated on 30/Apr/18 $$\mathrm{Stirling}\:\mathrm{approximation}:…
Question Number 99600 by bemath last updated on 22/Jun/20 Answered by mathmax by abdo last updated on 22/Jun/20 $$\mathrm{y}^{''} \:+\mathrm{3y}^{'} \:+\mathrm{2y}\:=\mathrm{sin}\left(\mathrm{e}^{\mathrm{x}} \right) \\ $$$$\left(\mathrm{he}\right)\rightarrow\mathrm{y}^{''} \:+\mathrm{3y}^{'}…
Question Number 165093 by Stanley last updated on 25/Jan/22 $$\overset{{lim}} {{x}}\rightarrow\mathrm{3}\left(\mathrm{2}{x}+\mathrm{3}{x}−\mathrm{4}\right) \\ $$$$\mathrm{FAILED}\:\mathrm{TO}\:\mathrm{CALCULATE} \\ $$ Commented by Ar Brandon last updated on 25/Jan/22 $$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\left(\mathrm{2}{x}+\mathrm{3}{x}−\mathrm{4}\right)…
Question Number 34007 by rahul 19 last updated on 29/Apr/18 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\sqrt{{x}−\mathrm{2}}\:+\sqrt{{x}}\:−\sqrt{\mathrm{2}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}\:\:{is}\:? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 29/Apr/18 $$={li}\underset{{x}\rightarrow\mathrm{2}} {{m}}\frac{\sqrt{{x}−\mathrm{2}\:}\:+\left({x}−\mathrm{2}\right)/\left(\sqrt{{x}}\:+\sqrt{\mathrm{2}}\right)}{\:\sqrt{\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)}\:} \\…
Question Number 99505 by bemath last updated on 21/Jun/20 Commented by Dwaipayan Shikari last updated on 21/Jun/20 $${Ans}:\mathrm{1} \\ $$ Commented by Dwaipayan Shikari last…
Question Number 33942 by manishprajapati076@gmail.com last updated on 28/Apr/18 Commented by manishprajapati076@gmail.com last updated on 28/Apr/18 $${solve}\mathrm{12}\:\mathrm{11}\:\mathrm{no}. \\ $$ Answered by MJS last updated on…
Question Number 164955 by atata01 last updated on 24/Jan/22 $$\mathrm{comment}\:\mathrm{creer}\:\mathrm{un}\:\mathrm{tableau}\:\mathrm{de}\:\mathrm{variation}\:\mathrm{a}\:\mathrm{partir}\:\mathrm{de}\:\mathrm{l}'\mathrm{application}? \\ $$ Commented by Ar Brandon last updated on 24/Jan/22 $$\begin{array}{|c|c|c|}{{x}}&\hline{−\pi\:\:\:\:\:\:\:\:\:\:\:−\frac{\pi}{\mathrm{2}}}&\hline{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}&\hline{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\pi}{\mathrm{2}}}&\hline{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\pi}\\{\mathrm{f}\:'\left({x}\right)}&\hline{\:\:\:\:\:\:\:\:\:\:\:\:−}&\hline{\:\:\:\:\:\:\:\:\:+}&\hline{\:\:\:\:\:\:\:\:\:\:+}&\hline{\:\:\:\:\:\:\:−}\\{\mathrm{f}\left({x}\right)}&\hline{\overset{\mathrm{0}} {\:}\:\:\:\:\:\:\:\:\:\searrow\:\:_{\:\:\:\:\:−\mathrm{1}} }&\hline{\:\:\:\:\:\:\:\:\:\nearrow^{\:\:\:\:\:\mathrm{0}} }&\hline{\underset{\mathrm{0}} {\:}\:\:\:\:\:\:\:\nearrow^{\:\:\:\:\:\:\mathrm{1}}…
Question Number 33865 by 33 last updated on 26/Apr/18 $${evaluate} \\ $$$$\:{li}\underset{{x}\rightarrow\infty} {{m}}\:\:\:\pi\:\frac{\left({a}\pi\right)^{{x}} }{{x}!} \\ $$ Commented by abdo imad last updated on 26/Apr/18 $${we}\:{have}\:{for}\:{x}\:\in{V}\left(+\infty\right)\:\:{x}!\:\sim\:{x}^{{x}}…
Question Number 99392 by AwaisAhmed last updated on 20/Jun/20 Commented by PRITHWISH SEN 2 last updated on 21/Jun/20 $$\mathrm{To}\:\mathrm{be}\:\mathrm{continuous}\: \\ $$$$\mathrm{4b}=−\mathrm{2}\Rightarrow\boldsymbol{\mathrm{b}}=\:−\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$ Answered by…