Question Number 82448 by zainal tanjung last updated on 21/Feb/20 $$\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{Lim}}\:\left\{\frac{\mathrm{sin}\:\left(\mathrm{6x}−\mathrm{3}\pi\right)^{\mathrm{2}} −\mathrm{sin}\:\left(\mathrm{6x}−\mathrm{3}\pi\right)\mathrm{sin}\:\left(\mathrm{4x}−\mathrm{2}\pi\right)}{\mathrm{5x}^{\mathrm{2}} \:\mathrm{cos}\:\left(\mathrm{5x}−\frac{\mathrm{5}\pi}{\mathrm{2}}\:\right)}\right. \\ $$ Commented by jagoll last updated on 21/Feb/20 $${let}\:{x}−\frac{\pi}{\mathrm{2}}\:=\:{u} \\…
Question Number 82426 by zainal tanjung last updated on 21/Feb/20 $$\mathrm{Lim}\:\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} }\:=\:… \\ $$$$\mathrm{x}\rightarrow\mathrm{0} \\ $$ Commented by jagoll last updated…
Question Number 82425 by zainal tanjung last updated on 21/Feb/20 $$\mathrm{Lim}\:\left(\frac{\mathrm{1}}{\mathrm{ex}}\right)^{\mathrm{6x}} =….. \\ $$$$\mathrm{x}\rightarrow\mathrm{0} \\ $$ Commented by abdomathmax last updated on 21/Feb/20 $${let}\:{f}\left({x}\right)=\left(\frac{\mathrm{1}}{{ex}}\right)^{\mathrm{6}{x}} \:\Rightarrow{f}\left({x}\right)={e}^{−\mathrm{6}{xln}\left({ex}\right)}…
Question Number 147862 by mathmax by abdo last updated on 24/Jul/21 $$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{n}} } \\ $$ Answered by qaz last updated on 24/Jul/21…
Question Number 147738 by liberty last updated on 23/Jul/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{arctan}\:{x}−\mathrm{arcsin}\:{x}}{\mathrm{sin}\:^{\mathrm{3}} {x}}\:=? \\ $$ Answered by gsk2684 last updated on 23/Jul/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}^{−\mathrm{1}} {x}−\mathrm{sin}^{−\mathrm{1}} {x}}{\mathrm{sin}\:^{\mathrm{3}}…
Question Number 82175 by jagoll last updated on 19/Feb/20 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{3}\:}]{\mathrm{8}{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}−\sqrt[{\mathrm{3}\:}]{\mathrm{8}{x}^{\mathrm{3}} +\mid{px}\mid^{\mathrm{2}} −\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${find}\:{p}\: \\ $$ Commented by mr W last updated…
Question Number 82160 by M±th+et£s last updated on 18/Feb/20 $${prove}\:{that} \\ $$$$\underset{{x}\rightarrow\infty} {{lim}}\:{n}^{\mathrm{2}} \:\sqrt{\left(\mathrm{1}−{cos}\left(\frac{\mathrm{1}}{{n}}\right)\sqrt{\left(\mathrm{1}−{cos}\frac{\mathrm{1}}{{n}}\right)\sqrt{\left(\mathrm{1}−{cos}\frac{\mathrm{1}}{{n}}\right)…}}\right.}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by MJS last updated on 18/Feb/20 $${w}=\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{1}}{{n}}\right)\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{1}}{{n}}\right)\sqrt{…}}} \\…
Question Number 147528 by alcohol last updated on 21/Jul/21 $${q}_{{n}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}{cos}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right) \\ $$$$\left.{i}\right)\:{study}\:{the}\:{variation}\:{of}\:{q}_{{n}} \\ $$$$\left.{ii}\right) \\ $$$$\:{show}\:{that}\:{cosx}=\frac{{sin}\mathrm{2}{x}}{\mathrm{2}{sinx}}\:,\:\forall{x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right] \\ $$$$\left.{iii}\right) \\ $$$${deduce}\:{that}\:{q}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}}…
Question Number 147459 by liberty last updated on 21/Jul/21 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\sqrt[{\mathrm{3}}]{\mathrm{8}+{x}^{\mathrm{3}} }−\mathrm{2}}{{x}^{\mathrm{2}} }\:=? \\ $$ Answered by mathmax by abdo last updated on 21/Jul/21…
Question Number 147412 by vvvv last updated on 20/Jul/21 Answered by puissant last updated on 21/Jul/21 $$={lim}_{{k}\rightarrow\infty} \frac{\mathrm{1}}{{k}}\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{{n}^{\mathrm{2}} +\mathrm{3}{nk}+\mathrm{9}{k}^{\mathrm{2}} {sin}\left(\frac{{n}}{{k}}\right)}{{k}^{\mathrm{2}} } \\ $$$$={lim}_{{k}\rightarrow\infty}…