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} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 62180 by Mikael last updated on 17/Jun/19 $$\underset{{x}\rightarrow\infty} {{lim}}\:\frac{{senx}}{{x}} \\ $$ Commented by maxmathsup by imad last updated on 17/Jun/19 $${if}\:{you}\:{mean}\:{sinx}\:\:{we}\:{have}\:\:\mid{sinx}\mid\leqslant\mathrm{1}\:\Rightarrow\mid\frac{{sinx}}{{x}}\mid\leqslant\frac{\mathrm{1}}{\mid{x}\mid}\:\:{for}\:{all}\:{x}\neq\mathrm{0}\:\:{but} \\ $$$${lim}_{{x}\rightarrow\infty}…
Question Number 127501 by slahadjb last updated on 30/Dec/20 $${prove}\:{the}\:{convergence}\:{of}\:\left(\alpha_{{n}} \right)_{{n}\:} {such}\:{that}\: \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\alpha_{{n}} }{{n}}\right)−\mathrm{2}\alpha_{{n}} +\mathrm{1}=\mathrm{0} \\ $$ Answered by mindispower last updated on…
Question Number 127459 by liberty last updated on 30/Dec/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{{x}}}}\:\right)^{{x}^{\mathrm{2}} } \:=?\: \\ $$ Answered by john_santu last updated on 30/Dec/20 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{x}}}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{x}}{\mathrm{2x}+\mathrm{1}}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{2x}}{\mathrm{2x}+\mathrm{1}}} \\…
Question Number 192969 by mustafazaheen last updated on 01/Jun/23 Commented by mustafazaheen last updated on 01/Jun/23 $$\mathrm{how}\:\mathrm{is}\:\mathrm{solution}? \\ $$ Answered by MM42 last updated on…
Question Number 127421 by Study last updated on 29/Dec/20 $${li}\underset{{x}\rightarrow−\infty} {{m}}\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{3}}}{\mathrm{10}{x}−\mathrm{3}}=??? \\ $$ Answered by ebi last updated on 29/Dec/20 $$\underset{{x}\rightarrow−\infty} {{lim}}\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{3}}}{\mathrm{10}{x}−\mathrm{3}} \\…