Question Number 173190 by mathlove last updated on 08/Jul/22 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}\left(\mathrm{2}−{e}^{{x}} \right)}{{x}+{lnx}}=? \\ $$ Answered by thfchristopher last updated on 08/Jul/22 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{−{e}^{{x}} }{\mathrm{2}−{e}^{{x}} }}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}…
Question Number 107598 by mathDivergent last updated on 11/Aug/20 $$\:\:\:\:\mathrm{Calculate}:\:\:\sqrt{\mathrm{1}\:+\:\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{3}\:+\:\sqrt{\mathrm{4}\:+\:\sqrt{\mathrm{5}\:+\:…}}}}} \\ $$ Commented by Her_Majesty last updated on 11/Aug/20 $$\approx\mathrm{1}.\mathrm{75793275662} \\ $$ Commented by Dwaipayan…
Question Number 173098 by mathlove last updated on 06/Jul/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 107550 by ajfour last updated on 11/Aug/20 $${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{2}^{{x}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}/{x}} \:\:=\:? \\ $$ Answered by hgrocks last updated on 11/Aug/20 Commented by ajfour…
Question Number 107536 by bemath last updated on 11/Aug/20 $$\:\spadesuit\mathcal{B}{e}\mathcal{M}{ath}\bigstar \\ $$$${Without}\:{L}'{Hopital}\:{and}\:{series}\: \\ $$$${find}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}−{x}}{{x}^{\mathrm{3}} } \\ $$ Commented by bobhans last updated on 12/Aug/20…
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…