Question Number 6990 by Tawakalitu. last updated on 04/Aug/16 Answered by sou1618 last updated on 05/Aug/16 $${L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({x}\right)^{\mathrm{1}/{x}} \\ $$$${ln}\left({L}\right)=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}}{ln}\left({x}\right) \\ $$$${when}\:\mathrm{0}<{x} \\ $$$$\:\:\:\:\mathrm{0}<\frac{\mathrm{1}}{{x}},…
Question Number 72495 by Tony Lin last updated on 29/Oct/19 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{lnx}}{{x}}\right)^{\frac{{lnx}}{{x}}} =? \\ $$ Commented by mathmax by abdo last updated on 29/Oct/19 $${let}\:{f}\left({x}\right)=\left(\frac{{lnx}}{{x}}\right)^{\frac{{lnx}}{{x}}}…
Question Number 6944 by Tawakalitu. last updated on 03/Aug/16 $${Prove}\:{that}: \\ $$$$ \\ $$$${e}^{{x}} =\:{Limit}\left({u}\:\rightarrow\:+\:\infty\right)\:\left(\mathrm{1}\:+\:\frac{{x}}{{u}}\right)^{{u}} \: \\ $$ Answered by sou1618 last updated on 03/Aug/16…
Question Number 6939 by sou1618 last updated on 03/Aug/16 $$\mathrm{please}\:\mathrm{solve}\:{L} \\ $$$$\mathrm{but}\:\mathrm{do}\:\mathrm{not}\:\mathrm{use}\:\mathrm{L}'\mathrm{Hopital}'\mathrm{s}\:\mathrm{rule}. \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{e}^{{x}} −\mathrm{1}}\right) \\ $$ Commented by Yozzii last updated on 03/Aug/16…
Question Number 72462 by 20190927 last updated on 29/Oct/19 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4x}\:}\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{3}} \mathrm{arctan}\left(\mathrm{x}^{\mathrm{5}} \right)} \\ $$ Commented by kaivan.ahmadi last updated on 29/Oct/19 $${lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4}{x}}\left(\mathrm{1}−\frac{{x}^{\mathrm{4}}…
Question Number 72398 by 20190927 last updated on 28/Oct/19 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{x}−\mathrm{x}^{\mathrm{2}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)\right] \\ $$ Commented by mathmax by abdo last updated on 28/Oct/19 $${let}\:{f}\left({x}\right)={x}−{x}^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\:\:{we}\:{have}\:{ln}^{'}…
Question Number 137902 by benjo_mathlover last updated on 08/Apr/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} \:\sqrt{{x}}}\:=? \\ $$ Answered by greg_ed last updated on 08/Apr/21 $$ \\ $$$$\boldsymbol{\mathrm{L}}'\boldsymbol{\mathrm{Hopital}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{rule}} \\…
Question Number 137820 by bramlexs22 last updated on 07/Apr/21 $$\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right)}{\left(\mathrm{1}+\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\left(\pi−\mathrm{2}{x}\right)^{\mathrm{3}} }\:? \\ $$ Answered by SLVR last updated on 07/Apr/21 Commented by SLVR last…
Question Number 72251 by aliesam last updated on 26/Oct/19 Commented by mathmax by abdo last updated on 26/Oct/19 $${let}\:{f}\left({x}\right)=\frac{{ln}\left(\mathrm{1}+{ax}+{bx}^{\mathrm{2}} \right)−{ln}\left(\mathrm{1}−{ax}+{bx}^{\mathrm{2}} \right)}{\mathrm{1}−{cosx}} \\ $$$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)=\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\mathrm{1}−{u}\:+{o}\left({u}\right)\:\left({u}\rightarrow\mathrm{0}\right)\:\Rightarrow \\…
Question Number 137773 by Ajadiazeemolamilekan last updated on 06/Apr/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{px}−\mathrm{cos}\:{qx}}{{x}^{\mathrm{2}} } \\ $$ Answered by bemath last updated on 06/Apr/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2sin}\:\left(\frac{{p}+{q}}{\mathrm{2}}\:{x}\right)\mathrm{sin}\:\left(\frac{{p}−{q}}{\mathrm{2}}\:{x}\right)}{{x}^{\mathrm{2}} } \\…