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Question Number 190182 by SLVR last updated on 29/Mar/23 $${Li}\underset{{x}\rightarrow\pi/\mathrm{2}} {{m}}\frac{{sinx}−{sinx}^{{sinx}} }{\mathrm{1}−{sinx}+{logsinx}} \\ $$$$\left.{a}\left.\right)\left.\mathrm{4}\left.\:\:\:\:\:\:\:\:{b}\right)\mathrm{2}\:\:\:\:\:\:\:\:\:\:{c}\right)\mathrm{1}/\mathrm{2}\:\:\:\:\:\:\:\:{d}\right){none} \\ $$ Commented by SLVR last updated on 29/Mar/23 $${kindly}\:{help}\:{me} \\…
Question Number 190178 by mnjuly1970 last updated on 29/Mar/23 $$ \\ $$$$\:\:{calculate} \\ $$$$ \\ $$$$\:\:\:\:\mathrm{lim}_{\:\mathrm{n}\rightarrow\infty} \frac{\:\Gamma\:\left(\:\frac{\:{n}+\mathrm{3}}{\mathrm{2}}\:\right)}{{n}^{\:\frac{\mathrm{3}}{\mathrm{2}}} .\Gamma\:\left(\frac{{n}}{\mathrm{2}}\:\right)}\:=\:? \\ $$ Commented by Frix last updated…
Question Number 190165 by TUN last updated on 29/Mar/23 $${S}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$=>\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}_{{n}} =¿ \\ $$ Answered by JDamian last updated…
Question Number 124555 by mnjuly1970 last updated on 04/Dec/20 Answered by mnjuly1970 last updated on 05/Dec/20 $${solution}: \\ $$$$\:\:{f}\::\left[{a},{b}\right]\rightarrow\mathbb{R}^{+} \cup\left\{\mathrm{0}\right\}\:{is}\:{continuous}\:: \\ $$$$\:\:\:\:{lim}_{{n}\rightarrow\infty} \left(\int_{{a}} ^{\:{b}} {f}^{\:{n}}…
Question Number 124548 by mnjuly1970 last updated on 04/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:\:\:\:\:\:\:{evaluate}\:::: \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \left\{\frac{\mathrm{1}}{{x}}\left[\frac{{ln}\left(\Gamma\left(\mathrm{1}+{x}\right)\right.}{{x}}−\psi\left({x}+\mathrm{1}\right)\right]\right\}=? \\ $$$$ \\ $$ Answered by mindispower last updated on…
Question Number 190082 by Rupesh123 last updated on 27/Mar/23 Answered by a.lgnaoui last updated on 29/Mar/23 $${L}=\frac{{e}^{{x}^{\mathrm{3}} } }{{x}^{\mathrm{3}} }−\frac{\mathrm{3}\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} +{x}\right)}}{\left({x}^{\mathrm{3}} −\mathrm{1}\right)+\mathrm{1}}+\frac{\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}} \\ $$$$\frac{{e}^{{x}}…
Question Number 124537 by nimnim last updated on 04/Dec/20 $${Please}\:{help} \\ $$$$ \\ $$$$\:\:{Show}\:{that}\:\:\:\:\:\:\underset{{n}\rightarrow\infty} {{lim}}\left(\frac{\left(\mathrm{3}{n}\right)!}{\left({n}!\right)^{\mathrm{3}} }\right)^{\mathrm{1}/{n}} =\:\mathrm{27} \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 124491 by Mammadli last updated on 03/Dec/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}^{{n}} }{{n}!}\:=\:\mathrm{0} \\ $$ Answered by mathmax by abdo last updated on 03/Dec/20 $$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{2}^{\mathrm{n}}…
Question Number 58948 by Mikael_Marshall last updated on 01/May/19 $$\underset{{x}\rightarrow+\infty} {{lim}}\:\frac{\sqrt{{x}}}{\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}} \\ $$ Commented by maxmathsup by imad last updated on 01/May/19 $$={lim}_{{x}\rightarrow+\infty} \:\:\:\:\frac{\sqrt{{x}}}{\:\sqrt{{x}}\sqrt{\mathrm{1}+\sqrt{\frac{{x}+\sqrt{{x}}}{{x}^{\mathrm{2}} }}}}\:={lim}_{{x}\rightarrow+}…