Question Number 128187 by john_santu last updated on 05/Jan/21 $$\:\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}}{\:\sqrt{\mathrm{1}−\mathrm{cos}\:{x}}}\:=?\: \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{e}^{\alpha{x}} −{e}^{\beta{x}} }{\mathrm{sin}\:\alpha{x}−\mathrm{sin}\:\beta{x}}=? \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 62626 by Tawa1 last updated on 23/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{of}\:\:\:\:\:\frac{\mathrm{n}!}{\mathrm{4}^{\mathrm{n}} }\:\:\:\mathrm{as}\:\:\mathrm{n}\:\:\mathrm{approach}\:\mathrm{infinity} \\ $$ Commented by mathmax by abdo last updated on 23/Jun/19 $${let}\:{u}_{{n}} =\frac{{n}!}{\mathrm{4}^{{n}} }\:\:\:\:{we}\:{have}\:{n}!\:\sim{n}^{{n}}…
Question Number 128149 by MathSh last updated on 04/Jan/21 $$\underset{{x}\rightarrow−\mathrm{1}} {{lim}}\left(\frac{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}}{{x}+\mathrm{2}−{x}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{{x}+\mathrm{1}}} =\:? \\ $$ Answered by liberty last updated on 04/Jan/21 $$\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\left(\frac{\mathrm{2x}^{\mathrm{3}}…
Question Number 62571 by Joel122 last updated on 23/Jun/19 $$\mathrm{Find} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left({n}\left(\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{3}\right)^{\mathrm{2}} }\:+\:…\:+\:\frac{\mathrm{1}}{\left(\mathrm{4}{n}−\mathrm{1}\right)^{\mathrm{2}} }\right)\right) \\ $$ Commented by Prithwish sen last updated on…
Question Number 62539 by aliesam last updated on 22/Jun/19 $$\mathrm{li}\underset{\mathrm{x}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{sin}\left(\mathrm{2x}\right)−\mathrm{2sin}\left(\mathrm{x}\right)}{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{3}} \right)−\left(\mathrm{arctan}\left(\mathrm{x}\right)\right)^{\mathrm{3}} } \\ $$ Answered by tanmay last updated on 22/Jun/19 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}}…
Question Number 127961 by bramlexs22 last updated on 03/Jan/21 $$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\sqrt[{\mathrm{n}}]{\mathrm{a}}\:+\:\sqrt[{\mathrm{n}}]{\mathrm{b}}}{\mathrm{2}}\:\right)^{\mathrm{n}} \:=\:?\: \\ $$$$\:\mathrm{a},\:\mathrm{b}\:\in\mathbb{R}\: \\ $$ Answered by liberty last updated on 03/Jan/21 $$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\sqrt[{\mathrm{n}}]{\mathrm{a}}\:+\sqrt[{\mathrm{n}}]{\mathrm{b}}}{\mathrm{2}}\right)^{\mathrm{n}}…
Question Number 62332 by tanmay last updated on 19/Jun/19 Answered by tanmay last updated on 20/Jun/19 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\sqrt{\mathrm{2}}\:×{n}^{{n}−\frac{\mathrm{1}}{\mathrm{2}}} }{{n}^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} ×\sqrt{\mathrm{2}\pi}\:×{e}^{−{n}} }×\left\{\frac{\left(\mathrm{2}×{n}^{\frac{\mathrm{1}}{{n}}} −\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\right\}^{\frac{{n}\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{{ln}^{\mathrm{2}} {n}}}…
Question Number 127846 by bemath last updated on 02/Jan/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{3}} \:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)}\:?\: \\ $$ Commented by Dwaipayan Shikari last updated on 02/Jan/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{xsin}\left(\frac{\mathrm{1}}{{x}}\right)}{{sin}^{\mathrm{2}}…
Question Number 62200 by maxmathsup by imad last updated on 17/Jun/19 $${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{ln}\left(\mathrm{1}+{x}+{sinx}\right)−{ln}\left(\mathrm{1}+{sin}\left(\mathrm{2}{x}\right)\right)}{{x}^{\mathrm{2}} } \\ $$ Commented by maxmathsup by imad last updated on 18/Jun/19…
Question Number 62192 by Sardor2211 last updated on 17/Jun/19 Answered by mr W last updated on 17/Jun/19 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}}…