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Question Number 89341 by Henri Boucatchou last updated on 17/Apr/20 $$\underset{{n}\rightarrow\infty} {{lim}}\sqrt[{{n}}]{\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)}=? \\ $$ Commented by Henri Boucatchou last updated on 17/Apr/20 $$\boldsymbol{{Check}}\:\:\boldsymbol{{once}}\:\:\boldsymbol{{more}}\:\:\boldsymbol{{Sir}},\:\:\:\boldsymbol{{ln}}\underset{{i}}…
Question Number 23777 by anoop7760@gmail.com last updated on 06/Nov/17 $${please}\:{solve}\:{question}\:{no}\:\mathrm{23764}\:{and}\:\mathrm{23765} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 23764 by Anoop kumar last updated on 05/Nov/17 $$\underset{{x}\rightarrow\propto} {\mathrm{lim}}\:\frac{\left(\mathrm{2}^{{x}^{{n}} } \overset{\mathrm{1}/{e}^{{x}} } {\right)}−\left(\mathrm{3}^{{x}^{{n}} } \right)^{\mathrm{1}/{e}^{{x}} } }{{x}^{{n}} } \\ $$$${where}\:{n}\varepsilon{N} \\ $$$$…
Question Number 23765 by Anoop kumar last updated on 05/Nov/17 $$\underset{{x}\rightarrow\propto} {\mathrm{lim}}\:\frac{\mathrm{cot}^{−\mathrm{1}} \left({x}^{−{a}} \mathrm{ln}\:_{{a}} \:{x}\right)}{\mathrm{sec}^{−\mathrm{1}} \left({a}^{{x}} \:\mathrm{ln}\:_{{x}} {a}\right)}\:\: \\ $$$$\left({a}>\mathrm{1}\right) \\ $$$$ \\ $$ Terms…
Question Number 154776 by john_santu last updated on 21/Sep/21 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}−\frac{\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}}\right)}{\frac{\mathrm{1}}{{x}}}}{{e}−\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} }\right)^{{x}} =? \\ $$ Commented by mathdanisur last updated on 21/Sep/21 $$\mathrm{say}\:\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{t}\:\Rightarrow\:\mathrm{t}\rightarrow\mathrm{0} \\ $$$$\mathrm{e}^{\underset{\boldsymbol{\mathrm{t}}\rightarrow\mathrm{0}}…
Question Number 23688 by A1B1C1D1 last updated on 04/Nov/17 Answered by $@ty@m last updated on 04/Nov/17 $${Use}\:{method}\:{of}\:{rationalization} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{{x}^{\mathrm{2}} \left(\sqrt{{x}^{\mathrm{2}} +\mathrm{12}}+\sqrt{\mathrm{12}}\right)}{{x}^{\mathrm{2}} +\mathrm{12}−\mathrm{12}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}}…
Question Number 154719 by liberty last updated on 21/Sep/21 $$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2ln}\:\left[\Gamma\left(\mathrm{sin}\:{x}\right)\right]−\mathrm{ln}\:\pi}{\Gamma\left(\mathrm{sec}\:{x}\right)−\mathrm{1}}\:=? \\ $$ Answered by john_santu last updated on 21/Sep/21 $$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:{x}\:\Psi^{\mathrm{0}} \left(\mathrm{sin}\:\left({x}\right)\right)}{\mathrm{2tan}\:\left(\mathrm{2}{x}\right)\mathrm{sec}\:\left(\mathrm{2}{x}\right)\Gamma\left(\mathrm{sec}\:\left(\mathrm{2}{x}\right)\right)\Psi^{\mathrm{0}} \left(\mathrm{sec}\:\left(\mathrm{2}{x}\right)\right)}= \\…
Question Number 23584 by A1B1C1D1 last updated on 02/Nov/17 Answered by FilupES last updated on 02/Nov/17 $$\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{3}+{x}\right)−\mathrm{ln}\left({x}\right)}{{x}}=\left(\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\right)\left(\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\left\{\mathrm{ln}\left(\mathrm{3}+{x}\right)−\mathrm{ln}\left({x}\right)\right\}\right) \\ $$$$\:…