Question Number 119867 by Study last updated on 27/Oct/20 $${li}\underset{{x}\rightarrow\infty} {{m}}\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)^{{n}} −\sqrt{{e}}}{\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)^{{n}} −{e}^{\mathrm{2}} }=??? \\ $$ Commented by Dwaipayan Shikari last updated on 27/Oct/20 $$\frac{\mathrm{1}}{\mathrm{16}}{e}^{−\frac{\mathrm{3}}{\mathrm{2}}}…
Question Number 54258 by 951172235v last updated on 01/Feb/19 Commented by Meritguide1234 last updated on 02/Feb/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 119725 by bemath last updated on 26/Oct/20 $$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\mathrm{4}\sqrt{\mathrm{2}}−\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)^{\mathrm{5}} }{\left(\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\:=? \\ $$ Commented by bemath last updated on 26/Oct/20 $${compare}\:{via}\:{L}'{Hopital} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}}…
Question Number 185209 by Rupesh123 last updated on 18/Jan/23 Answered by som(math1967) last updated on 18/Jan/23 $$\left.{a}\right)\:\underset{{n}\rightarrow\infty} {{lim}}\:\frac{\mathrm{1}}{{n}}\left[\:\left(\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2022}} +\left(\frac{\mathrm{2}}{{n}}\right)^{\mathrm{2022}} +…\left(\frac{{n}}{{n}}\right)^{\mathrm{2022}} \right] \\ $$$$\underset{{n}\rightarrow\infty} {{lim}}\:\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}}…
Question Number 119636 by I want to learn more last updated on 25/Oct/20 $$\mathrm{Test}\:\mathrm{wheather}\:\mathrm{convergent} \\ $$$$\:\:\:\underset{\mathrm{n}\:\:=\:\:\mathrm{1}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{3n}\:\:−\:\:\mathrm{1}}{\mathrm{7}^{\mathrm{n}} \:\:+\:\:\mathrm{2n}} \\ $$ Answered by 1549442205PVT last…
Question Number 119629 by 9764954060 last updated on 25/Oct/20 Answered by TANMAY PANACEA last updated on 25/Oct/20 $${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$${so}\:{here}\:{x}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{3}^{\mathrm{2}} −\mathrm{4}×\mathrm{2}×\mathrm{7}}}{\mathrm{2}×\mathrm{2}}=\frac{−\mathrm{3}\pm\sqrt{−\mathrm{47}}}{\mathrm{4}}…
Question Number 185106 by mathlove last updated on 17/Jan/23 Commented by mathlove last updated on 17/Jan/23 $${answer}\:{this}\:{problum}??? \\ $$ Commented by Frix last updated on…
Question Number 119517 by bemath last updated on 25/Oct/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\mathrm{sin}\:\left({x}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{sin}\:{x}\:\right]\:=? \\ $$ Answered by Dwaipayan Shikari last updated on 25/Oct/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({sin}\left({x}+\frac{\mathrm{1}}{{x}}\right)−{sinx}\right) \\ $$$$=\underset{{x}\rightarrow\infty}…
Question Number 119482 by liberty last updated on 24/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}^{\mathrm{2}} \:\mathrm{cos}\:\left(\frac{\mathrm{1}}{{x}}\right)\:? \\ $$ Answered by Olaf last updated on 25/Oct/20 $$\forall\:{x}\in\mathbb{R}^{\ast} ,\:−\mathrm{1}\:\leqslant\:\mathrm{cos}\left(\frac{\mathrm{1}}{{x}}\right)\:\leqslant\:+\mathrm{1} \\ $$$$\forall\:{x}\in\mathbb{R}^{\ast}…
Question Number 184960 by mathlove last updated on 14/Jan/23 Answered by cortano1 last updated on 15/Jan/23 $${L}=−\frac{\mathrm{9}}{\mathrm{4}}\left(\mathrm{ln}\:\mathrm{4}−\mathrm{1}\right) \\ $$ Commented by mathlove last updated on…