Question Number 173795 by Ar Brandon last updated on 18/Jul/22 $$\:\:\:\:\mathrm{Soient}\:{a}\:\mathrm{et}\:{b}\:\mathrm{deux}\:\mathrm{r}\acute {\mathrm{e}els}.\:\mathrm{Pour}\:\mathrm{tout}\:{n}\:\in\:\mathbb{N}\:\mathrm{on}\:\mathrm{pose} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{u}_{{n}} =\sqrt{{n}}+{a}\sqrt{{n}+\mathrm{1}}+{b}\sqrt{{n}+\mathrm{2}}. \\ $$$$\:\:\mathrm{1}.\:\:\mathrm{V}\acute {\mathrm{e}rifier}\:\mathrm{que}\:\mathrm{la}\:\mathrm{suite}\:\left({u}_{{n}} \right)\:\mathrm{tend}\:\mathrm{vers}\:\mathrm{0}\:\mathrm{si}\:\mathrm{et}\:\mathrm{seulement}\:\mathrm{si}\:{a}+{b}=−\mathrm{1}. \\ $$$$\:\:\mathrm{2}.\:\:\mathrm{D}\acute {\mathrm{e}terminer}\:{a}\:\mathrm{et}\:{b}\:\mathrm{pour}\:\mathrm{que}\:\mathrm{la}\:\mathrm{s}\acute {\mathrm{e}rie}\:\Sigma{u}_{{n}} \:\mathrm{soit}\:\mathrm{convergente}.\:\: \\…
Question Number 42713 by Rio Michael last updated on 01/Sep/18 Commented by Rio Michael last updated on 01/Sep/18 $${find}\:{the}\:{Area}\:{bounded}\:{by}\:{the}\:{curve}\:{y}=\:{x}^{\mathrm{2}} ,{the}\:{x}−{axis}\:{and}\: \\ $$$${the}\:{line}\:{x}=\mathrm{2}. \\ $$ Answered…
Question Number 108228 by bemath last updated on 15/Aug/20 $$\:\:\frac{\curlyvee\mathcal{B}{e}\mathcal{M}{ath}\curlyvee}{\pitchfork} \\ $$$$\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{n}+\mathrm{ln}\:{a}}{{n}}\right)^{\frac{{n}}{{b}}} ?\: \\ $$ Answered by Dwaipayan Shikari last updated on 15/Aug/20 $$\underset{{n}\rightarrow\infty}…
Question Number 108217 by john santu last updated on 15/Aug/20 $$\:\:\:\:\frac{\heartsuit{JS}\heartsuit}{\leqslant°\equiv°\leqslant} \\ $$$$\:\underset{{x}\rightarrow\pi/\mathrm{6}} {\mathrm{lim}}\frac{\mathrm{2}−\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{\left(\mathrm{6}{x}−\pi\right)^{\mathrm{2}} }\:?\: \\ $$ Commented by john santu last updated on 15/Aug/20…
Question Number 173729 by a.lgnaoui last updated on 17/Jul/22 $${Evaluate} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{{x}−\frac{\mathrm{1}}{\:\sqrt{{x}}−{x}}+\mathrm{1}}{{x}^{\mathrm{2}} +\frac{\sqrt{{x}}}{\:\sqrt{{x}}−\mathrm{1}}−\mathrm{1}} \\ $$ Answered by blackmamba last updated on 17/Jul/22 $$\:\:{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{{x}−\frac{\mathrm{1}}{\:\sqrt{{x}}−{x}}\:−\mathrm{1}}{{x}^{\mathrm{2}}…
Question Number 173734 by mnjuly1970 last updated on 17/Jul/22 $$ \\ $$$$\:\mathrm{lim}_{\:{n}\rightarrow\infty} \:\left({n}\:+\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{x}^{\:{n}} \:\mathrm{ln}\left(\:\mathrm{1}+{x}\:\right){dx}=? \\ $$$$ \\ $$ Answered by mindispower last updated…
Question Number 42636 by bohemian last updated on 30/Aug/18 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{r}}{{n}^{\mathrm{2}} +{n}+{r}} \\ $$$$ \\ $$ Commented by bohemian last updated on 30/Aug/18…
Question Number 107975 by john santu last updated on 13/Aug/20 $$\:\:\:\:\:\:\frac{\clubsuit{JS}\heartsuit}{\bullet\equiv\bullet} \\ $$$$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right)−\mathrm{tan}\:\left(\mathrm{sin}\:{x}\right)}{{x}−\mathrm{sin}\:{x}\:} \\ $$$$\:\left(\mathrm{2}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} \sqrt{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right)\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right)\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right)\sqrt{…}}}}\:?\: \\ $$$$ \\ $$ Commented by malwaan…
Question Number 173498 by blackmamba last updated on 12/Jul/22 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{8cot}\:\left({x}\right)+\mathrm{9}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}}\right)}{\mathrm{12}\:\mathrm{csc}\:\left({x}\right)−\mathrm{4sin}\:\left(\frac{\mathrm{1}}{{x}}\right)}\:=? \\ $$ Commented by blackmamba last updated on 13/Jul/22 $$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{8cos}\:{x}\left(\frac{{x}}{\mathrm{sin}\:{x}}\right)+\mathrm{9}{x}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}}\right)}{\frac{\mathrm{12}{x}}{\mathrm{sin}\:{x}}−\mathrm{4}{x}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$$\:=\:\frac{\mathrm{8}+\mathrm{0}}{\mathrm{12}−\mathrm{0}}\:=\:\frac{\mathrm{8}}{\mathrm{12}}=\frac{\mathrm{2}}{\mathrm{3}} \\…
Question Number 173499 by cortano1 last updated on 12/Jul/22 Commented by blackmamba last updated on 12/Jul/22 $$\:{x}=\mathrm{1}+{r}\: \\ $$$$\:{L}\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{7}+\left(\mathrm{1}+{r}\right)^{\mathrm{3}} }−\sqrt{\mathrm{3}+\left(\mathrm{1}+{r}\right)^{\mathrm{2}} }}{{r}} \\ $$$$\:\:\:\:\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{8}+\left(\mathrm{3}{r}+\mathrm{3}{r}^{\mathrm{2}}…