Question Number 162081 by cortano last updated on 26/Dec/21 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\sqrt{\frac{\mathrm{1}}{{x}}+\sqrt{\frac{\mathrm{1}}{{x}}+\sqrt{\frac{\mathrm{1}}{{x}}}}}\:−\sqrt{\frac{\mathrm{1}}{{x}}−\sqrt{\frac{\mathrm{1}}{{x}}+\sqrt{\frac{\mathrm{1}}{{x}}}}}\:=?\right. \\ $$ Answered by mr W last updated on 26/Dec/21 $$\frac{\mathrm{1}}{{L}}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{x}}+\sqrt{\frac{\mathrm{1}}{{x}}+\sqrt{\frac{\mathrm{1}}{{x}}}}}\:−\sqrt{\frac{\mathrm{1}}{{x}}−\sqrt{\frac{\mathrm{1}}{{x}}+\sqrt{\frac{\mathrm{1}}{{x}}}}}\:} \\ $$$$\:=\underset{{x}\rightarrow\mathrm{0}}…
Question Number 96489 by Ar Brandon last updated on 01/Jun/20 $$\mathcal{G}\mathrm{iven}\:\:\begin{cases}{\mathrm{u}_{\mathrm{0}} =\mathrm{a}}\\{\mathrm{u}_{\mathrm{n}+\mathrm{1}} =\mathrm{a}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{n}} }\right)\mathrm{u}_{\mathrm{n}} }\end{cases}\:\:\mathrm{a}>\mathrm{0}\:\:\mathrm{b}>\mathrm{1} \\ $$$$\mathrm{a}\backslash\:\:\mathcal{C}\mathrm{alculate}\:\:\mathrm{u}_{\mathrm{1}} ,\:\:\mathrm{u}_{\mathrm{2}} ,\:\:\mathrm{and}\:\:\mathrm{u}_{\mathrm{3}} . \\ $$$$\mathrm{b}\backslash\:\:\mathcal{S}\mathrm{how}\:\:\mathrm{that}\:\:\mathrm{the}\:\:\mathrm{sequence}\:\:\left(\mathrm{u}_{\mathrm{n}} \right)_{\mathrm{n}\in\mathbb{N}} \:\:\mathrm{is}\:\:\mathrm{increasing}. \\…
Question Number 161867 by cortano last updated on 23/Dec/21 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{9}}\:−\sqrt{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{4}}\:\right)^{\mathrm{4}{x}} \:=\:? \\ $$ Answered by Ar Brandon last updated on 23/Dec/21 $$\mathcal{A}=\underset{{x}\rightarrow\infty}…
Question Number 96319 by M±th+et+s last updated on 31/May/20 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} }{{e}}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$ Answered by abdomathmax last updated on 31/May/20 $$\mathrm{f}\left(\mathrm{x}\right)=\left(\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} }{\mathrm{e}}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:\Rightarrow\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} }{\mathrm{e}}\right)…
Question Number 30758 by abdo imad last updated on 25/Feb/18 $${find}\:{lim}_{{n}\rightarrow\infty} \left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+\frac{\mathrm{1}}{{n}+\mathrm{2}}\:+….+\frac{\mathrm{1}}{{n}+{p}}\right)\:{pfixed}\:{fromN}^{\bigstar} \\ $$$$ \\ $$ Commented by abdo imad last updated on 28/Feb/18 $${let}\:{put}\:{u}_{{n}}…
Question Number 30759 by abdo imad last updated on 25/Feb/18 $${find}\:{lim}_{{n}\rightarrow\infty} \:\:\:\:\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} \:. \\ $$ Commented by abdo imad last updated on 27/Feb/18 $${let}\:{use}\:{the}\:{stirling}\:{formula}\:{we}\:{have}…
Question Number 30757 by abdo imad last updated on 25/Feb/18 $${find}\:{lim}_{{n}\rightarrow\infty} \:\:\left(\frac{\mathrm{1}}{{n}}\:+\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} \:−\mathrm{1}}}\:+….\:+\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} \:−\left({n}−\mathrm{1}\right)^{\mathrm{2}} }}\:\right) \\ $$ Commented by abdo imad last updated on 28/Feb/18…
Question Number 30755 by abdo imad last updated on 25/Feb/18 $${find}\:\:\:{lim}_{{n}\rightarrow\infty} \:\:^{{n}} \sqrt{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)…\left(\mathrm{1}+\frac{{n}}{{n}}\right)\:} \\ $$ Commented by abdo imad last updated on 28/Feb/18 $${let}\:{put}\:{A}_{{n}} =^{{n}}…
Question Number 161801 by qaz last updated on 22/Dec/21 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{n}} ^{\mathrm{n}\left(\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{n}} } \left(\mathrm{1}−\frac{\mathrm{4}}{\mathrm{x}}\right)^{\mathrm{x}} \mathrm{dx}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 96232 by joki last updated on 30/May/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{cos}\:\sqrt{{x}}−\mathrm{cos}\:{x}}{\mathrm{1}−\mathrm{cos}\:\sqrt{{x}}}= \\ $$ Commented by bobhans last updated on 31/May/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\sqrt{\mathrm{x}}−\mathrm{cos}\:\mathrm{x}}{\mathrm{1}−\mathrm{cos}\:\sqrt{\mathrm{x}}\:}\:=\:\mathrm{does}\:\mathrm{not}\:\mathrm{exist}\:! \\ $$ Answered…