Question Number 162287 by qaz last updated on 28/Dec/21 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt[{\mathrm{n}}]{\mathrm{x}^{\mathrm{n}} +\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{n}} }\mathrm{dx}=? \\ $$ Commented by mr W last updated on 28/Dec/21…
Question Number 96684 by Ar Brandon last updated on 03/Jun/20 $$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{n}+\mathrm{k}}{\mathrm{n}^{\mathrm{2}} +\mathrm{k}^{\mathrm{2}} } \\ $$$$\left\{\mathrm{Reimann}'\mathrm{s}\:\:\mathrm{integral}\:\:\mathrm{may}\:\:\mathrm{help}\right\} \\ $$ Answered by Sourav mridha last…
Question Number 96685 by Ar Brandon last updated on 03/Jun/20 $$\underset{\omega\rightarrow\infty} {\mathrm{lim}20log}\sqrt{\mathrm{1}+\left(\frac{\omega}{\mathrm{100}}\right)^{\mathrm{2}} } \\ $$ Answered by Rio Michael last updated on 04/Jun/20 $$\mathrm{By}\:\mathrm{l}'\mathrm{hopitals}\:\mathrm{rule}\:\mathrm{i}\:\mathrm{get}\:+\infty\:\mathrm{as}\:\mathrm{answer} \\…
Question Number 162182 by bobhans last updated on 27/Dec/21 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{2}}\:\right)^{\mathrm{2022}} +\left(\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{5}}\:\right)^{\mathrm{2022}} }{\mathrm{x}^{\mathrm{2022}} }\:=\:? \\ $$ Answered by cortano last updated on 27/Dec/21…
Question Number 96602 by Ar Brandon last updated on 03/Jun/20 $$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}}}\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\sqrt{\mathrm{k}} \\ $$ Answered by Sourav mridha last updated on 03/Jun/20 $$\boldsymbol{{by}}\:\boldsymbol{{inspection}}\:\underset{\boldsymbol{{k}}=\mathrm{1}}…
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}}…