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Author: Tinku Tara

lim-n-2-n-4-n-3-2-n-3-4n-

Question Number 131885 by Study last updated on 09/Feb/21 $${lim}_{{n}\rightarrow−\infty} \frac{\mathrm{2}^{{n}} +\mathrm{4}^{{n}−\mathrm{3}} }{\mathrm{2}^{{n}−\mathrm{3}} +\mathrm{4}{n}}=? \\ $$ Commented by malwan last updated on 09/Feb/21 $$\underset{{n}\rightarrow−\infty} {{lim}}\:\frac{\mathrm{2}^{{n}}…

log-2-x-log-3-x-1-x-

Question Number 131884 by Study last updated on 09/Feb/21 $${log}_{\mathrm{2}} {x}+{log}_{\mathrm{3}} {x}=\mathrm{1}\:\:\:\:\:\:\:{x}=? \\ $$ Answered by EDWIN88 last updated on 09/Feb/21 $$\:\frac{\mathrm{ln}\:\mathrm{x}}{\mathrm{ln}\:\mathrm{2}}\:+\:\frac{\mathrm{ln}\:\mathrm{x}}{\mathrm{ln}\:\mathrm{3}}\:=\:\mathrm{1}\: \\ $$$$\:\mathrm{ln}\:\mathrm{x}\:\left(\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{3}}\right)=\mathrm{1} \\…

let-I-n-0-e-nt-1-e-t-n-1-dt-n-from-N-prove-the-existence-of-I-n-2-find-lim-n-I-n-

Question Number 66351 by mathmax by abdo last updated on 12/Aug/19 $${let}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{{nt}} }{\left(\mathrm{1}+{e}^{{t}} \right)^{{n}+\mathrm{1}} }{dt}\:\:\:\:\:\left({n}\:{from}\:{N}^{\bigstar} \right) \\ $$$$\left.\right){prove}\:{the}\:{existence}\:{of}\:{I}_{{n}} \\ $$$$\left.\mathrm{2}\right){find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:{I}_{{n}} \\…

Question-131887

Question Number 131887 by Algoritm last updated on 09/Feb/21 Answered by SEKRET last updated on 09/Feb/21 $$\:\boldsymbol{\mathrm{Leybnist}}\:\:\:\boldsymbol{\mathrm{formula}} \\ $$$$\:\:\boldsymbol{\mathrm{u}}=\:\boldsymbol{\mathrm{e}}^{−\mathrm{2}\boldsymbol{\mathrm{x}}} \:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{u}}^{\boldsymbol{\mathrm{n}}} =\left(−\mathrm{2}\right)^{\boldsymbol{\mathrm{n}}} \centerdot\boldsymbol{\mathrm{e}}^{−\mathrm{2}\boldsymbol{\mathrm{x}}} \\ $$$$\:\boldsymbol{\mathrm{v}}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{x}}}}\:\:\:\:\:\:\boldsymbol{\mathrm{v}}^{\boldsymbol{\mathrm{n}}} =\frac{\left(\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}\right)!!}{\mathrm{2}^{\boldsymbol{\mathrm{n}}}…