Question Number 41998 by Joel578 last updated on 16/Aug/18 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}\:+\:\mathrm{cos}\:{x}}{{x}\:+\:\mathrm{sin}\:{x}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}+\frac{{cosx}}{{x}}}{\mathrm{1}+\frac{{sinx}}{{x}}} \\ $$$${the}\:{value}\:{of}\:{sinx}\:{lies}\:{in}\:{between}\:\pm\mathrm{1}\:{whatever} \\…
Question Number 107516 by bemath last updated on 11/Aug/20 $$\:\:\:\:\circledcirc\mathcal{B}{e}\mathcal{M}{ath}\circledcirc \\ $$$$\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{cos}\:\mathrm{2}{t}}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−{t}\right)−\mathrm{cos}\:\mathrm{2}{t}}\:? \\ $$ Answered by bemath last updated on 11/Aug/20 $$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{t}}{\mathrm{cos}\:{t}−\mathrm{cos}\:\mathrm{2}{t}}\:×\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{cos}\:\mathrm{2}{t}}}…
Question Number 173053 by mathlove last updated on 06/Jul/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 107500 by Ar Brandon last updated on 11/Aug/20 $$\mathrm{Given}\:\mathrm{a}\:\in\mathbb{R}−\left\{\pm\mathrm{1}\right\} \\ $$$$\mathrm{1}.\:\mathrm{Show}\:\mathrm{that}\:\forall\mathrm{x}\in\mathbb{R}\:\mathrm{1}−\mathrm{2acos}\left(\mathrm{x}\right)+\mathrm{a}^{\mathrm{2}} >\mathrm{0} \\ $$$$\mathrm{2}.\:\mathrm{Show}\:\mathrm{that}; \\ $$$$\:\:\:\:\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{1}−\mathrm{2acos}\left(\frac{\mathrm{2k}\pi}{\mathrm{n}}\right)+\mathrm{a}^{\mathrm{2}} \right)=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{a}−\mathrm{e}^{\mathrm{2ik}\pi/\mathrm{n}} \right)\left(\mathrm{a}−\mathrm{e}^{−\mathrm{2ik}\pi/\mathrm{n}} \right)…
Question Number 107498 by Ar Brandon last updated on 11/Aug/20 $$\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}}…