Question Number 109728 by bemath last updated on 25/Aug/20 $$\:\:\:\bigtriangleup\frac{\flat\epsilon}{{math}}\bigtriangledown \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:\mathrm{3}{x}+\mathrm{1}} \\ $$ Commented by bemath last updated on 25/Aug/20 $${thank}\:{you}\:{all}\:{master} \\…
Question Number 109722 by bemath last updated on 25/Aug/20 $$\:\:\:\:\:\multimap\frac{\flat\epsilon}{{math}}\multimap \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{1}−\mathrm{cos}\:{x}}}\:=\:?\: \\ $$ Commented by bemath last updated on 25/Aug/20 Answered by mathmax…
Question Number 175217 by mnjuly1970 last updated on 23/Aug/22 $$ \\ $$$$\:\:{lim}_{\:{x}\rightarrow\mathrm{0}} \left\{\:\lfloor\:\frac{{cos}\left({x}\right)}{{x}}\:\rfloor\:−\lfloor\frac{\mathrm{1}+{cos}\left({x}\right)}{\mathrm{1}−{cos}\left({x}\right)}\:\rfloor\right\} \\ $$ Commented by kaivan.ahmadi last updated on 24/Aug/22 $${we}\:{know}\:{that}\:\left[{x}\right]_{{x}\rightarrow\infty} \sim{x}\:{so} \\…
Question Number 109639 by Ar Brandon last updated on 24/Aug/20 Answered by Dwaipayan Shikari last updated on 25/Aug/20 $${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }=\left(−\mathrm{1}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}}…
Question Number 44085 by rk last updated on 21/Sep/18 $${lim}\:\mathrm{x}\rightarrow\mathrm{0}\:\left[\frac{\mathrm{sin}\:\mid{x}\mid}{{x}}\right] \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 22/Sep/18 Commented by tanmay.chaudhury50@gmail.com last updated on…
Question Number 175147 by infinityaction last updated on 20/Aug/22 $$\:\:\:\mathrm{if}\:\mathrm{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:\mathrm{in}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\:\:\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\:\:\:\underset{{m}\rightarrow\infty\:} {\mathrm{lim}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\mathrm{1}+\mathrm{cos}^{\mathrm{2}{m}} \left({n}!\pi{x}\right)\right]\: \\ $$ Answered by floor(10²Eta[1]) last updated on…
Question Number 175137 by cortano1 last updated on 20/Aug/22 $$\:\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{5}}]{\mathrm{5}{x}−\mathrm{9}}\:\sqrt[{\mathrm{4}}]{\mathrm{4}{x}−\mathrm{7}}\:\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−\mathrm{5}}\:\sqrt{\mathrm{2}{x}−\mathrm{3}}−\mathrm{1}}{{x}−\mathrm{2}}=? \\ $$ Answered by Ar Brandon last updated on 20/Aug/22 $$\mathscr{L}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{5}}]{\mathrm{5}{x}−\mathrm{9}}\sqrt[{\mathrm{4}}]{\mathrm{4}{x}−\mathrm{7}}\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−\mathrm{5}}\sqrt{\mathrm{2}{x}−\mathrm{3}}−\mathrm{1}}{{x}−\mathrm{2}} \\ $$$$\:\:\:\:\:=\underset{{t}\rightarrow\mathrm{0}}…
Question Number 175115 by mnjuly1970 last updated on 19/Aug/22 Answered by aleks041103 last updated on 19/Aug/22 $${n}\rightarrow\infty \\ $$$$\sqrt{\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }}−\mathrm{1}=\left(\mathrm{1}+\frac{{k}}{\mathrm{2}{n}^{\mathrm{2}} }\right)−\mathrm{1}=\frac{{k}}{\mathrm{2}{n}^{\mathrm{2}} } \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}}\:\underset{{k}=\mathrm{1}}…
Question Number 109462 by 150505R last updated on 23/Aug/20 Answered by mathmax by abdo last updated on 24/Aug/20 $$\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left[\mathrm{k}^{\mathrm{2}} \mathrm{x}\:+\mathrm{k}^{\mathrm{2}} \right]\:\Rightarrow\:\mathrm{A}_{\mathrm{n}}…
Question Number 174951 by infinityaction last updated on 14/Aug/22 $$\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\left[\mathrm{1}+\sqrt{\mathrm{2}}+^{\mathrm{3}} \sqrt{\mathrm{3}}+…^{{n}} \sqrt{{n}}\right] \\ $$ Answered by TheHoneyCat last updated on 16/Aug/22 $$\mathrm{There}'\mathrm{s}\:\mathrm{a}\:\mathrm{theorem}\:\mathrm{that}\:\mathrm{sates}\:\mathrm{that} \\ $$$${U}_{{n}}…