Question Number 213241 by RoseAli last updated on 01/Nov/24 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}−\mathrm{tan}\:{x}}{{x}^{\mathrm{3}} } \\ $$ Answered by ajfour last updated on 01/Nov/24 $$=−\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{sin}\:{x}}{{x}}×\left(\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{4}\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)×\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\right\} \\…
Question Number 213169 by MrGaster last updated on 31/Oct/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{prove} \\ $$$$\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{n}}{\mathrm{1}+{n}^{\mathrm{2}} {x}^{\mathrm{2}} }{e}^{{x}^{\mathrm{2}} } {dx}=\frac{\pi}{\mathrm{2}}. \\ $$ Answered by Berbere last…
Question Number 213082 by liuxinnan last updated on 30/Oct/24 Answered by MrGaster last updated on 30/Oct/24 $$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}^{\frac{\mathrm{1}}{{n}}} \approx\int_{\mathrm{1}} ^{{n}} {x}^{\frac{\mathrm{1}}{{n}}} {dx} \\ $$$$\int{x}^{\frac{\mathrm{1}}{{x}}}…
Question Number 213109 by mathlove last updated on 30/Oct/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}{\:\sqrt{{x}+\mathrm{1}}}=? \\ $$ Answered by MrGaster last updated on 30/Oct/24 $$=\frac{\sqrt{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{{x}}\right)}}{\:\sqrt{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)}} \\ $$$$=\frac{\sqrt{{x}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{{x}}}}{\:\sqrt{{x}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}} \\ $$$$=\frac{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{{x}}}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}}…
Question Number 213073 by ajfour last updated on 29/Oct/24 $${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}−\mathrm{1}\right)\right\} \\ $$ Answered by universe last updated on 29/Oct/24 $${L}\:=\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 212984 by Faetmaaa last updated on 27/Oct/24 $$\underline{\boldsymbol{\mathrm{Notation}}\::}\:\mathrm{Soit}\:{A}\:\mathrm{une}\:\mathrm{partie}\:\mathrm{de}\:\mathbb{R}.\:\mathrm{On}\:\mathrm{appelle}\:{indicatrice}\:{de}\:{A}, \\ $$$$\mathrm{not}\acute {\mathrm{e}e}\:\chi_{{A}} ,\:\mathrm{l}'\mathrm{application}\:{x}\: \:\begin{cases}{\mathrm{1}\:\mathrm{si}\:{x}\:\in\:{A}}\\{\mathrm{0}\:\mathrm{sinon}}\end{cases}. \\ $$$$ \\ $$$$\mathrm{1}.\:\mathrm{Pour}\:{k}\:\mathrm{dans}\:\mathbb{N}^{\ast} \:\mathrm{notons}\:{f}_{{k}} \::\:{x}\: \:\left(\mathrm{cos}\:{x}\right)^{\mathrm{2}{k}} . \\ $$$$\mathrm{Montrer}\:\mathrm{que}\:\left({f}_{{k}}…
Question Number 212808 by RoseAli last updated on 24/Oct/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\left({x}^{.\mathrm{1}} −{x}^{.\mathrm{9}} \right)^{\mathrm{1}{o}} −{x}\right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212762 by universe last updated on 23/Oct/24 Commented by MrGaster last updated on 23/Oct/24 $$\mathrm{Rewrite}\:\mathrm{the}\:\mathrm{integrand}\:\mathrm{function}\:\mathrm{as}: \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right)=\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right) \\ $$$$\mathrm{Computational}\:\mathrm{integral}: \\ $$$$\int_{\mathrm{0}} ^{{n}}…
Question Number 212745 by RoseAli last updated on 22/Oct/24 Answered by mehdee7396 last updated on 22/Oct/24 $${lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}+{tanx}}{\mathrm{1}−{tanx}}−\mathrm{1}\right)×\frac{\mathrm{1}}{{sinx}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{\mathrm{2}{tanx}}{\mathrm{1}−{tanx}}\right)×\frac{\mathrm{1}}{{sinx}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{\mathrm{2}}{\mathrm{1}−{tanx}}\right)×\frac{\mathrm{1}}{{cosx}}=\mathrm{2} \\…
Question Number 212495 by liuxinnan last updated on 15/Oct/24 Terms of Service Privacy Policy Contact: info@tinkutara.com