Question Number 164530 by hoochhoch last updated on 18/Jan/22 Commented by Rasheed.Sindhi last updated on 18/Jan/22 $$\mathrm{sin}\alpha=−\frac{\mathrm{12}}{\mathrm{3}}=−\mathrm{4}\Rightarrow\alpha\notin\mathbb{R}\:{in}\:{Q}#\mathrm{1} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 98986 by Ar Brandon last updated on 17/Jun/20 $$\mathrm{Is}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{u}_{\mathrm{n}} =\mathrm{cos}\left(\frac{\mathrm{n}\pi}{\mathrm{20}}\right)\:\mathrm{divergent}? \\ $$ Answered by maths mind last updated on 17/Jun/20 $${u}_{\mathrm{40}{n}+\mathrm{10}} =\mathrm{0} \\…
Question Number 98952 by Ar Brandon last updated on 17/Jun/20 $$\mathrm{Without}\:\mathrm{using}\:\mathrm{L}'\mathrm{H}\hat {\mathrm{o}pital}'\mathrm{s}\:\mathrm{rule}\:\mathrm{or}\:\mathrm{Maclaurin}'\mathrm{s}\:\mathrm{expansion} \\ $$$$\mathrm{series},\:\mathrm{find}\:\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{xe}^{\mathrm{x}} }{\mathrm{e}^{\mathrm{x}} −\mathrm{1}}−\mathrm{1}}{\mathrm{x}} \\ $$ Commented by 675480065 last updated on…
Question Number 98955 by Ar Brandon last updated on 17/Jun/20 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\centerdot\centerdot\centerdot}}}} \\ $$ Commented by mr W last updated on 17/Jun/20 $$=\:{Q}\mathrm{98858} \\ $$…
Question Number 98938 by john santu last updated on 17/Jun/20 Answered by mathmax by abdo last updated on 17/Jun/20 $$=\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{n}^{\mathrm{2}} \left(\mathrm{3}+\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\right)\left(\mathrm{2}+\frac{\mathrm{5}}{\mathrm{n}}\right)}{\mathrm{n}^{\mathrm{2}} \left(\mathrm{7}+\frac{\mathrm{2}}{\mathrm{n}}\right)\left(\mathrm{5}−\frac{\mathrm{cosn}}{\mathrm{n}}\right)}\:=\frac{\mathrm{3}×\mathrm{2}}{\mathrm{7}×\mathrm{5}}\:=\frac{\mathrm{6}}{\mathrm{35}} \\…
Question Number 98880 by Ar Brandon last updated on 16/Jun/20 $$\mathcal{G}\mathrm{iven}\:\mathrm{the}\:\mathrm{sequence}\:\left(\mathrm{u}_{\mathrm{n}} \right)_{\mathrm{n}\in\mathbb{N}^{\ast} } \:\mathrm{defined}\:\mathrm{by}\:\begin{cases}{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\:\mathrm{if}\:\mathrm{n}\equiv\mathrm{0mod}\left(\mathrm{3}\right)}\\{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{n}} }+\mathrm{1}\:\:\:\mathrm{if}\:\mathrm{n}\equiv\mathrm{1mod}\left(\mathrm{3}\right)}\\{\frac{\mathrm{u}_{\mathrm{n}−\mathrm{1}} +\mathrm{u}_{\mathrm{n}+\mathrm{2}} }{\mathrm{2}}\:\mathrm{if}\:\mathrm{n}\equiv\mathrm{2mod}\left(\mathrm{3}\right)}\end{cases} \\ $$$$\mathrm{a}\backslash\mathcal{D}\mathrm{etermine}\:\mathrm{the}\:\mathrm{first}−\mathrm{8}^{\mathrm{th}} \:\mathrm{terms}\:\mathrm{of}\:\left(\mathrm{u}_{\mathrm{n}} \right)_{\mathrm{n}\in\mathbb{N}^{\ast} } \\ $$$$\mathrm{b}\backslash\mathcal{S}\mathrm{how}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequences}\:\left(\mathrm{v}_{\mathrm{n}}…
Question Number 98858 by M±th+et+s last updated on 16/Jun/20 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}…….}}}} \\ $$$$ \\ $$ Commented by Tinku Tara last updated on 17/Jun/20 $$\mathrm{Hi}\:\mathrm{David} \\…
Question Number 33313 by abdo imad last updated on 14/Apr/18 $${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{{ln}\left(\mathrm{1}+{sinx}\right)\:−{sin}\left({ln}\left(\mathrm{1}+{x}\right)\right)}{{x}^{\mathrm{2}} } \\ $$ Commented by abdo imad last updated on 17/Apr/18 $${let}\:{put}\:{u}\left({x}\right)={ln}\left(\mathrm{1}+{sinx}\right)\:−{sin}\left({ln}\left(\mathrm{1}+{x}\right)\right){and}\:{v}\left({x}\right)={x}^{\mathrm{2}} \\…
Question Number 33312 by abdo imad last updated on 14/Apr/18 $${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{e}^{−\mathrm{3}{x}^{\mathrm{2}} } \:−\mathrm{1}}{{x}^{\mathrm{2}} }\:. \\ $$ Answered by Joel578 last updated on 14/Apr/18 $$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 98833 by john santu last updated on 16/Jun/20 Commented by bobhans last updated on 16/Jun/20 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{3}^{\mathrm{n}} \:\left(\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{n}} }\right)\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \right)\:=\:\mathrm{3}\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{n}} }\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \\…