Question Number 157123 by cortano last updated on 20/Oct/21 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{log}\:_{\mathrm{sin}\:{x}} \left(\mathrm{cos}\:{x}\right)}{\mathrm{log}\:_{\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)} \left(\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\right)}=? \\ $$ Commented by john_santu last updated on 21/Oct/21 $${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)\:\mathrm{ln}\:\left(\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\right)}{\mathrm{ln}\:\left(\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\right).\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)} \\…
Question Number 26020 by abdo imad last updated on 17/Dec/17 $${answer}\:{to}\:\mathrm{25962}…{we}\:{S}=\sum_{{n}=\mathrm{1}} ^{\propto} \:\mathrm{1}/_{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)} \:{and}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\mathrm{1}/_{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)} \\ $$$${we}\:{have}\:{S}=\:{lim}_{{n}−>\propto} \:{S}_{{n}} \:{we}\:{decompose}\:{the}\:{the}\:{rational}\:{fraction} \\ $$$${F}\left({X}\right)=\:\:\:\mathrm{1}/_{{X}}…
Question Number 157055 by cortano last updated on 19/Oct/21 Commented by MathSh last updated on 20/Oct/21 $$=\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\boldsymbol{\pi}} {\mathrm{lim}}\frac{\pi^{\boldsymbol{\mathrm{x}}^{\boldsymbol{\pi}} } \centerdot\pi\mathrm{x}^{\boldsymbol{\mathrm{x}}-\mathrm{1}} \centerdot\mathrm{ln}\left(\pi\right)-\mathrm{x}^{\boldsymbol{\pi}^{\boldsymbol{\mathrm{x}}} } \left(\pi^{\boldsymbol{\mathrm{x}}} \centerdot\mathrm{ln}\left(\pi\right)\centerdot\mathrm{ln}\left(\mathrm{x}\right)+\pi^{\boldsymbol{\mathrm{x}}} \centerdot\frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{1}}…
Question Number 25962 by abdo imad last updated on 16/Dec/17 $${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{{n}=\propto} \:\:\:\mathrm{1}/_{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)} \:\:{we}\:{give}\:\:\:\sum_{{n}=\mathrm{1}} ^{{n}=\propto} \mathrm{1}/_{{n}} \mathrm{2}=\:\pi^{\mathrm{2}} /\mathrm{6} \\ $$$${and}\:\:{H}_{{n}} =\mathrm{1}+\mathrm{2}^{−\mathrm{1}} +\mathrm{3}^{−\mathrm{1}} +…+{n}^{−\mathrm{1}} =\:{ln}\left({n}\right)\:+\:{s}\:+\:\theta\left(\mathrm{1}/{n}\right)\:…
Question Number 156993 by amin96 last updated on 18/Oct/21 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{n}} {\underbrace{\left({sin}\left({sin}\left({sin}\ldots\left({sin}\left({x}\right)\right)\ldots\right)}}\:\sqrt{{n}}=?\right.\right. \\ $$$$\mathrm{0}<{x}<\pi \\ $$ Commented by MathSh last updated on 18/Oct/21 $$\mathrm{0}<{x}<\pi\:\:\mathrm{or}\:\:\mathrm{0}\leqslant{x}\leqslant\pi.? \\…
Question Number 91419 by john santu last updated on 30/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{4}} {x}}\left(\mathrm{sin}\:\left(\frac{{x}}{\mathrm{1}+{x}}\right)−\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}\right) \\ $$ Commented by john santu last updated on 01/May/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{4}}…
Question Number 25852 by abdo imad last updated on 15/Dec/17 $${let}\:{s}\:{put}\:{H}_{{n}} \:=\:\mathrm{1}\:+\mathrm{2}^{−\mathrm{1}} +\mathrm{3}^{−\mathrm{1}} +….+{n}^{−\mathrm{1}} \:{and}\:\:\:{U}_{{n}} =\:{H}_{{n}} \:−{ln}\left({n}\right) \\ $$$$\:{prove}\:{that}\:{U}_{{n}} \:{is}\:{convergent}\:{to}\:{a}\:{number}\:\:{s}\:{wish}\:{verify} \\ $$$$\mathrm{0}<{s}<\mathrm{1}\:\:\:\left({s}\:{is}\:{named}\:{number}\:{of}\:{Euler}\:\right) \\ $$ Commented…
Question Number 156891 by cortano last updated on 16/Oct/21 Answered by mindispower last updated on 17/Oct/21 $$\mathrm{1}−{tg}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}^{{k}} }\right)=\frac{{cos}\left(\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }\right)}{{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}^{{k}} }\right)} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}}…
Question Number 91274 by M±th+et+s last updated on 29/Apr/20 $$\underset{{n}\rightarrow\infty} {{lim}n}^{−{n}^{\mathrm{2}} } \left[\left({n}+\mathrm{1}\right)\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)….\left({n}+\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\right)\right]^{{n}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 156788 by mathlove last updated on 15/Oct/21 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…..{x}^{{n}} −{n}}{{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…..{x}^{{m}} −{m}}=?\:\:\:\:\frac{\mathrm{0}}{\mathrm{0}} \\ $$ Answered by puissant last updated on…