Question Number 73117 by aliesam last updated on 06/Nov/19 Commented by mathmax by abdo last updated on 06/Nov/19 $${we}\:{have}\:{cos}\left(\mathrm{3}{x}\right)\sim\mathrm{1}−\frac{\left(\mathrm{3}{x}\right)^{\mathrm{2}} }{\mathrm{2}}\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow{cos}\left(\mathrm{3}{x}\right)−\mathrm{1}\:\sim−\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{1}−{cos}\left(\mathrm{3}{x}\right)\sim\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\frac{\mathrm{1}−{cos}\left(\mathrm{3}{x}\right)}{{x}^{\mathrm{2}} }\sim\frac{\mathrm{9}}{\mathrm{2}}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}}…
Question Number 138603 by Ar Brandon last updated on 15/Apr/21 $$\frac{\mathrm{1}}{\mathrm{2}\pi\mathrm{i}}\underset{\mathrm{T}\rightarrow\infty} {\mathrm{lim}}\underset{\gamma−\mathrm{iT}} {\overset{\gamma+\mathrm{iT}} {\int}}\frac{\mathrm{e}^{\mathrm{st}} }{\mathrm{s}−\mathrm{a}}\mathrm{ds} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 138576 by liberty last updated on 15/Apr/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{cot}\:\mathrm{2}{x}\right)^{\mathrm{2tan}\:\mathrm{2}{x}} \:=?\: \\ $$ Answered by phanphuoc last updated on 15/Apr/21 $${li}\underset{{u}\left({x}\right)−>\mathrm{0}} {{m}}\left(\mathrm{1}+{u}\left({x}\right)\right)^{\mathrm{1}/{u}\left({x}\right)} ={e} \\…
Question Number 7501 by Yozzia last updated on 01/Sep/16 $${Compute}\:\underset{{A}\rightarrow+\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{A}}\int_{\mathrm{1}} ^{{A}} {A}^{\frac{\mathrm{1}}{{x}}} {dx}\right\}. \\ $$$$\left({IMC}\:\mathrm{2015}\right) \\ $$ Answered by FilupSmith last updated on 01/Sep/16…
Question Number 138571 by bemath last updated on 15/Apr/21 $$\begin{cases}{{u}_{\mathrm{1}} =\mathrm{1}}\\{{u}_{{n}+\mathrm{1}} =\:\frac{{n}^{\mathrm{2}} −{n}+\mathrm{1}}{{n}^{\mathrm{2}} }}\end{cases}\:;\:\forall{n}\in\mathbb{R} \\ $$$$\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{u}_{{n}} \:=? \\ $$ Answered by mathmax by abdo…
Question Number 138473 by henderson last updated on 14/Apr/21 $$\boldsymbol{\mathrm{hi}}\:! \\ $$$$\boldsymbol{\mathrm{calculate}}\::\:\underset{{x}\rightarrow\infty} {{lim}}\:\left[\frac{{ln}\left({x}+\mathrm{1}\right)}{{ln}\:\left({x}\right)}\right]^{{x}\:{ln}\:\left({x}\right)} . \\ $$ Answered by mathmax by abdo last updated on 14/Apr/21…
Question Number 138437 by EnterUsername last updated on 13/Apr/21 $$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}\right)!!}=\sqrt{\mathrm{e}} \\ $$ Answered by Ar Brandon last updated on 13/Apr/21 $$\:\:\:\:\:\left(\mathrm{2n}\right)!!=\mathrm{2n}\left(\mathrm{2n}−\mathrm{2}\right)\left(\mathrm{2n}−\mathrm{4}\right)…\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}^{\mathrm{n}}…
Question Number 72905 by Tony Lin last updated on 04/Nov/19 $${f}\left({x}\right)\geqslant\mathrm{0},\:{and}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{0},\underset{{x}\rightarrow{a}} {\mathrm{lim}}{g}\left({x}\right)=\infty \\ $$$${then}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)^{{g}\left({x}\right)} =? \\ $$ Commented by mathmax by abdo last…
Question Number 7363 by Yozzia last updated on 25/Aug/16 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{ln}\left({n}\right)}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{tan}^{−\mathrm{1}} {i}}{{n}+\mathrm{1}−{i}}\right)\right\}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 72836 by aliesam last updated on 03/Nov/19 Commented by mathmax by abdo last updated on 04/Nov/19 $${changement}\:{x}=\frac{\pi}{\mathrm{3}}+{t}\:\:{give} \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{3}}} \:\:\frac{\sqrt{\mathrm{3}}{cosx}−{sinx}}{\mathrm{1}−\mathrm{2}{cosx}}\:={lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\frac{\sqrt{\mathrm{3}}{cos}\left(\frac{\pi}{\mathrm{3}}+{t}\right)−{sin}\left(\frac{\pi}{\mathrm{3}}+{t}\right)}{\mathrm{1}−\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}+{t}\right)} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}}…