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 $$\mathrm{There}'\mathrm{s}\:\mathrm{a}\:\mathrm{theorem}\:\mathrm{that}\:\mathrm{sates}\:\mathrm{that} \\ $$$${U}_{{n}}…
Question Number 174915 by cortano1 last updated on 14/Aug/22 $$\:{Find}\:{the}\:{value}\:{of}\:{a}\:{for}\:{which}\: \\ $$$$\:{the}\:{limit}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({ax}\right)−\mathrm{arctan}\:{x}−{x}}{{x}^{\mathrm{3}} +{x}^{\mathrm{4}} } \\ $$$${is}\:{finite}\:{and}\:{then}\:{evaluate}\:{the}\:{limit} \\ $$ Answered by Ar Brandon last updated…
Question Number 174883 by infinityaction last updated on 13/Aug/22 $$\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{x}^{\mathrm{2}} } \mathrm{sin}\left({nx}\right){dx}\: \\ $$ Answered by Mathspace last updated on 14/Aug/22 $${u}_{{n}}…
Question Number 174876 by infinityaction last updated on 13/Aug/22 $$\:\:\mathrm{let}\:{f}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\mathbb{R}\:{be}\:{a}\:\mathrm{continuous} \\ $$$$\:\mathrm{function}\:\mathrm{ditermine}\:\left(\mathrm{with}\:\mathrm{appropriate}\right. \\ $$$$\left.\mathrm{justification}\right)\:\mathrm{the}\:\mathrm{following}\: \\ $$$$\:\:\mathrm{limit}:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}} {f}\left({x}\right){dx} \\ $$ Answered by TheHoneyCat…
Question Number 174863 by cortano1 last updated on 13/Aug/22 Commented by kaivan.ahmadi last updated on 13/Aug/22 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{\mathrm{1}−\left(\mathrm{1}+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\right)\left(\mathrm{1}−\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\right)}{{x}^{\mathrm{8}} }= \\ $$$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{\mathrm{1}−\left(\mathrm{1}−\frac{{x}^{\mathrm{8}} }{\mathrm{4}}\right)}{{x}^{\mathrm{8}}…
Question Number 174841 by infinityaction last updated on 12/Aug/22 $$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\mathrm{sin}\left(\mathrm{sin}{x}\right)−\mathrm{sin}^{\mathrm{2}} {x}\:\:}{{x}^{\mathrm{6}} } \\ $$ Answered by Ar Brandon last updated on 12/Aug/22 $$\mathcal{A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\mathrm{sin}\left(\mathrm{sin}{x}\right)−\mathrm{sin}^{\mathrm{2}}…