Question Number 107498 by Ar Brandon last updated on 11/Aug/20 $$\mathrm{Show}\:\mathrm{that}\: \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{a}−\mathrm{e}^{\frac{\mathrm{2}{i}\mathrm{k}\pi}{\mathrm{n}}} \right)\left(\mathrm{a}−\mathrm{e}^{−\frac{\mathrm{2}{i}\mathrm{k}\pi}{\mathrm{n}}} \right)=\left(\mathrm{a}^{\mathrm{n}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$ Answered by 1549442205PVT last updated…
Question Number 107414 by bemath last updated on 10/Aug/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\circledcirc\mathcal{B}{e}\mathbb{M}{ath}\circledcirc \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}−\mathrm{3cos}\:^{\mathrm{6}} {x}\:\mathrm{cos}\:^{\mathrm{4}} \mathrm{2}{x}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{4}{x}+\mathrm{cos}\:{x}}{\mathrm{36}{x}^{\mathrm{2}} }\:\:\: \\ $$$$ \\ $$ Answered by ajfour last…
Question Number 172913 by cortano1 last updated on 03/Jul/22 $$\:\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\:\right)^{{x}} \:=? \\ $$ Answered by FongXD last updated on 03/Jul/22 $$=\underset{\mathrm{x}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\mathrm{2x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}}…
Question Number 107372 by bobhans last updated on 10/Aug/20 $$\:\:\:\gtrdot\boldsymbol{\mathcal{B}\mathrm{obhans}}\lessdot \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}\:\sqrt{\frac{\mathrm{2x}−\mathrm{cos}\:\mathrm{5x}}{\mathrm{3x}^{\mathrm{3}} }\:}\:? \\ $$ Answered by john santu last updated on 10/Aug/20 $$\:\:\:\:\:\:\:\:\divideontimes\mathcal{JS}\divideontimes…
Question Number 107377 by ajfour last updated on 10/Aug/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left(\mathrm{1}+{a}\mathrm{cos}\:{x}\right)−{b}\mathrm{sin}\:{x}}{{x}^{\mathrm{3}} }\:=\:\mathrm{1} \\ $$$${Find}\:\:{a}\:{and}\:{b}\:. \\ $$ Answered by john santu last updated on 10/Aug/20 $$\:\:\:\:\divideontimes\mathcal{JS}\divideontimes…
Question Number 107310 by bemath last updated on 10/Aug/20 Answered by bobhans last updated on 10/Aug/20 $$\:\:\:\:\:\:\:\curlyvee\boldsymbol{\mathrm{bobhans}}\curlyvee \\ $$$$\mathrm{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\mathrm{x}^{\mathrm{2}} \right)−\mathrm{1}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{2}}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}−\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{2}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\mathrm{x}^{\mathrm{2}}…
Question Number 107299 by bobhans last updated on 10/Aug/20 $$\:\:\:\:\:\:\maltese\mathrm{bobhans}\maltese \\ $$$$\mathrm{find}\:\mathrm{without}\:\mathrm{L}'\mathrm{Hopital}\:\mathrm{and}\:\mathrm{series}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{3}} }\:? \\ $$ Answered by john santu last updated on…
Question Number 41761 by math khazana by abdo last updated on 12/Aug/18 $${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:{x}\:{ln}\left(\mathrm{1}−{e}^{{sinx}} \right)\: \\ $$ Answered by alex041103 last updated on 12/Aug/18 $${As}\:{x}\rightarrow\mathrm{0}\:{the}\:{expression}\:{becomes}\:{one}…
Question Number 172766 by mathlove last updated on 01/Jul/22 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}!!−\sqrt{\frac{\mathrm{2}}{\pi}}}{\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\centerdot\centerdot\centerdot\right)−\Psi_{\mathrm{1}} \left({x}+\mathrm{1}\right)}=? \\ $$$${solve}\:{this}\:{pleas} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 107222 by bobhans last updated on 09/Aug/20 Answered by john santu last updated on 09/Aug/20 $$\:\:\:\:\:\:\:\boxdot\mathrm{JS}\boxdot \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{7}^{\mathrm{2x}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{x}} }\right)\right)^{\frac{\mathrm{2}}{\mathrm{x}}} =\:\mathrm{7}^{\mathrm{4}} \:×\underset{{x}\rightarrow\infty}…