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…
Question Number 99355 by bemath last updated on 20/Jun/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:−\mathrm{cot}\:^{\mathrm{2}} \mathrm{x}\:\right)\:=? \\ $$ Commented by john santu last updated on 20/Jun/20 $$\mathrm{maybe}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}}…
Question Number 99350 by bemath last updated on 20/Jun/20 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{3}−\mathrm{x}}{\:\sqrt{\mathrm{9x}^{\mathrm{2}} −\mathrm{8}}\:}\:? \\ $$ Commented by bobhans last updated on 20/Jun/20 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{−\mathrm{x}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{x}}\right)}{−\mathrm{x}\sqrt{\mathrm{9}−\frac{\mathrm{8}}{\mathrm{x}^{\mathrm{2}} }}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\…
Question Number 99348 by bobhans last updated on 20/Jun/20 Commented by bemath last updated on 20/Jun/20 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{3x}+\sqrt{\mathrm{x}^{\mathrm{2}} }}{\mathrm{7x}−\mathrm{5}\sqrt{\mathrm{x}^{\mathrm{2}} }}\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{3x}−\mathrm{x}}{\mathrm{7x}+\mathrm{5x}} \\ $$$$=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{2x}}{\mathrm{12x}}\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\…
Question Number 99344 by bemath last updated on 20/Jun/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}^{\mathrm{x}} \:+\:\mathrm{3}^{\mathrm{x}} \:\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:? \\ $$ Answered by abdomsup last updated on 20/Jun/20 $${let}\:{f}\left({x}\right)=\left(\mathrm{2}^{{x}} \:+\mathrm{3}^{{x}}…
Question Number 99341 by bemath last updated on 20/Jun/20 $$\mathrm{if}\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\mathrm{x}−\sqrt{\mathrm{ax}+\mathrm{b}}}{\mathrm{2x}−\mathrm{6}}\:=\:−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\mathrm{then}\:\mathrm{a}−\mathrm{2b}\:=\: \\ $$ Commented by bobhans last updated on 20/Jun/20 $$\left(\mathrm{1}\right)\:\mathrm{3}−\sqrt{\mathrm{3a}+\mathrm{b}}\:=\:\mathrm{0}\: \\ $$$$\mathrm{9}\:=\:\mathrm{3a}+\mathrm{b}\:\Rightarrow\mathrm{b}\:=\:\mathrm{9}−\mathrm{3a}…