Question Number 226914 by Estevao last updated on 18/Dec/25 Answered by breniam last updated on 19/Dec/25 $$\mathrm{We}\:\mathrm{can}\:\mathrm{concider}\:\mathrm{it}\:\mathrm{only}\:\mathrm{for}\:\mathrm{odd}\:{n},\:\mathrm{because} \\ $$$$\mathrm{real}\:\mathrm{even}\:\mathrm{root}\:\mathrm{of}\:\mathrm{negative}\:\mathrm{number}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist}. \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{2}{n}−\mathrm{1}}]{\left(\mathrm{2}{n}−\mathrm{2}\right)!\left(\mathrm{2}−\mathrm{2}{n}\right)}= \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{2}{n}−\mathrm{1}}]{\left(\mathrm{2}{n}−\mathrm{1}\right)!}×\sqrt[{\mathrm{2}{n}−\mathrm{1}}]{\frac{\mathrm{2}−\mathrm{2}{n}}{\mathrm{2}{n}−\mathrm{1}}}…
Question Number 226877 by Spillover last updated on 17/Dec/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 226878 by Spillover last updated on 17/Dec/25 Answered by gregori last updated on 18/Dec/25 $$\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}}−\sqrt{{x}+{h}}}{\left(\sqrt{{x}+{h}}\:.\sqrt{{x}}\:\right){h}} \\ $$$$\:=\:\frac{\mathrm{1}}{{x}}\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−{h}}{\left(\sqrt{{x}}+\sqrt{{x}+{h}}\:\right){h}} \\ $$$$\:=\:−\frac{\mathrm{1}}{{x}.\:\mathrm{2}\sqrt{{x}}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{{x}}} \\ $$…
Question Number 226879 by Spillover last updated on 17/Dec/25 Answered by gregori last updated on 18/Dec/25 $$\left({a}\right)\:{f}\:'\left({x}\right)=\:−\mathrm{sin}\:{x}\:.\:{e}^{\mathrm{cos}\:{x}} \: \\ $$$$\:\left({b}\right)\:{f}\:'\left({x}\right)\:=\:\mathrm{3sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:{x}\:.{e}^{\mathrm{sin}\:^{\mathrm{3}} {x}} \\ $$$$ \\…
Question Number 226866 by Estevao last updated on 17/Dec/25 Commented by Jyrgen last updated on 17/Dec/25 $$\int\sqrt{\mathrm{1}+{x}^{{n}} }{dx}={x}\:_{\mathrm{1}} {F}_{\mathrm{2}} \:\left(−\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{{n}};\:\mathrm{1}+\frac{\mathrm{1}}{{n}};\:−{x}^{{n}} \right) \\ $$ Commented by…
Question Number 226860 by Estevao last updated on 17/Dec/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 226861 by Estevao last updated on 17/Dec/25 Answered by Estevao last updated on 17/Dec/25 $${Go}\:{go}\:{gooooooo} \\ $$ Answered by Tawa11 last updated on…
Question Number 226855 by Kassista last updated on 17/Dec/25 Answered by mehdee7396 last updated on 17/Dec/25 $${x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}=\mathrm{4}\Rightarrow{x}=\mathrm{0},\frac{\mathrm{8}}{\mathrm{5}} \\ $$$$\Rightarrow{P}\left(\mathrm{0},,\mathrm{2}\right)\:\&\:\:{Q}\left(\frac{\mathrm{8}}{\mathrm{5}},−\frac{\mathrm{6}}{\mathrm{5}}\right)\Rightarrow{m}_{{PQ}} =−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{tan}\alpha=\frac{\mathrm{3}}{\mathrm{4}}\backsimeq\mathrm{36}^{\mathrm{0}} \Rightarrow\angle{POQ}\backsimeq\mathrm{126}^{\mathrm{0}}…
Question Number 226880 by Spillover last updated on 17/Dec/25 Answered by TonyCWX last updated on 18/Dec/25 $$\int_{\mathrm{0}} ^{\mathrm{1}} \left[\mathrm{2}^{{x}} \right]\mathrm{d}{x} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left[{e}^{{x}\mathrm{ln}\left(\mathrm{2}\right)} \right]\mathrm{d}{x}…
Question Number 226882 by hardmath last updated on 17/Dec/25 Answered by breniam last updated on 17/Dec/25 $$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}}{\:\frac{\sqrt[{{n}}]{\mathrm{2}^{{n}+\mathrm{1}} }−\mathrm{1}}{\:\sqrt[{{n}}]{\mathrm{2}}−\mathrm{1}}}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}\left(\sqrt[{{n}}]{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}\sqrt[{{n}}]{\mathrm{2}}−\mathrm{1}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}\sqrt[{{n}}]{\mathrm{2}}−\mathrm{1}\right)=\mathrm{2}−\mathrm{1}=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\infty}…