Question Number 198572 by cortano12 last updated on 22/Oct/23 $$\:\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2n}}} {\mathrm{lim}}\:\frac{\sqrt{\frac{\mathrm{sin}\:\mathrm{2nx}}{\mathrm{1}+\mathrm{cos}\:\mathrm{nx}}}}{\mathrm{4n}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} −\pi^{\mathrm{2}} }\:=? \\ $$ Answered by MM42 last updated on 22/Oct/23 $$\frac{\pi}{\mathrm{2}{n}}−{x}={u} \\…
Question Number 198254 by mathocean1 last updated on 15/Oct/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 198152 by universe last updated on 11/Oct/23 $$\:\:\:\:{a}_{{n}+\mathrm{2}} \:=\:\:\:\sqrt{{a}_{{n}} ×{a}_{{n}+\mathrm{1}} }\:\:\:\forall\:{n}\geqslant\mathrm{1}\:,\:{n}\:\in\:\mathrm{N} \\ $$$$\:\mathrm{and}\:\mathrm{here}\:\:\mathrm{a}_{\mathrm{1}\:} =\:\alpha\:\:{and}\:{a}_{\mathrm{2}} =\:\beta\:\:\mathrm{then} \\ $$$$\:\:\:\mathrm{prove}\:\mathrm{that}\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}+\mathrm{2}} \:\:\:=\:\:\left(\alpha×\beta^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} \\ $$ Answered…
Question Number 198032 by cortano12 last updated on 08/Oct/23 $$\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{4025x}}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}+\mathrm{1}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}+\mathrm{2}}\:}\:+…\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{4025x}}}\:\right)=? \\ $$ Commented by universe last updated on 08/Oct/23 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{4025}}}×\frac{\mathrm{1}}{{x}}\underset{{r}=\mathrm{0}} {\overset{\mathrm{4024}{x}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{r}/{x}}} \\…
Question Number 197998 by mathlove last updated on 07/Oct/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{3}} }=? \\ $$$${with}\:{out}\:{L}'{Hospital}\:{rul} \\ $$ Answered by witcher3 last updated on 07/Oct/23 $$\mathrm{x}=\mathrm{sh}\left(\mathrm{t}\right),\mathrm{t}\rightarrow\mathrm{0}…
Question Number 197947 by mnjuly1970 last updated on 05/Oct/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{calculate}\ldots \\ $$$$\:\:\mathrm{L}\:=\:\mathrm{lim}\:_{\mathrm{n}\rightarrow\infty} \sqrt[{{n}}]{\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\ldots\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}\:=\:?\:\:\:\:\:\:\:\: \\ $$$$\:\: \\ $$ Answered by MM42 last updated on…
Question Number 197891 by Erico last updated on 02/Oct/23 $$\mathrm{Soit}\:\mathrm{I}=\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \sqrt{\mathrm{t}\sqrt{\mathrm{t}\left(\mathrm{1}−\mathrm{t}\right)}}\mathrm{dt} \\ $$$$\mathrm{Comment}\:\mathrm{calculer}\:\mathrm{I} \\ $$ Answered by Mathspace last updated on 03/Oct/23 $${I}=\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 197832 by mathlove last updated on 30/Sep/23 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}+\sqrt{{cosx}}}{−\mathrm{1}+\sqrt{{cosx}}}=? \\ $$ Answered by MM42 last updated on 30/Sep/23 $${let}\:\:{f}\left({x}\right)=\frac{\mathrm{2}+\sqrt{{cosx}}}{−\mathrm{1}+\sqrt{{cosx}}} \\ $$$${for}\:\:{a}_{{n}} =\mathrm{2}{n}\pi\Rightarrow{lim}_{{n}\rightarrow\infty} \:{f}\left({a}_{{n}}…
Question Number 197766 by universe last updated on 28/Sep/23 $$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\mathrm{sin}^{\mathrm{2}{n}+\mathrm{1}} {x}\:\mathrm{cos}\:{x}\right){dx}\:\:=\:? \\ $$ Commented by witcher3 last updated on 28/Sep/23 $$\mathrm{dominate}\:\mathrm{cv}\:\mathrm{Theorem} \\…
Question Number 197661 by pticantor last updated on 25/Sep/23 $$\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{m}}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}+….+{t}^{{n}} }=?\: \\ $$ Commented by JDamian last updated on 25/Sep/23 $$\mathrm{Why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{post}\:\mathrm{again}\:\mathrm{the}\:\mathrm{same}\:\mathrm{topic} \\…