Question Number 93695 by i jagooll last updated on 14/May/20 Commented by mathmax by abdo last updated on 14/May/20 $${f}\left({x}\right)\:=\frac{\mathrm{6}{x}−^{\mathrm{3}} \sqrt{\mathrm{27}{x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{1}}}{\left(^{\mathrm{3}} \sqrt{\mathrm{8}{x}^{\mathrm{3}} −{x}}\:+{x}\right)}\:\Rightarrow{f}\left({x}\right)\:=\frac{\mathrm{6}{x}−\mathrm{3}{x}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{27}{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{27}{x}^{\mathrm{3}}…
Question Number 159229 by tounghoungko last updated on 14/Nov/21 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}+{x}−\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}}\:\right)=? \\ $$ Answered by FongXD last updated on 14/Nov/21 $$\mathrm{L}=\underset{\mathrm{x}\rightarrow+\infty} {\mathrm{lim}x}\left(\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}}−\mathrm{x}+\mathrm{2x}−\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}}…
Question Number 28151 by tawa tawa last updated on 21/Jan/18 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\left(\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{e}} \mathrm{x}}\:−\:\frac{\mathrm{x}}{\mathrm{x}\:−\:\mathrm{1}}\right) \\ $$ Commented by çhëý böý last updated on 21/Jan/18 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{\left({x}−\mathrm{1}\right)−{xlnx}}{\left({x}−\mathrm{1}\right){lnx}}\right)=\frac{\mathrm{0}}{\mathrm{0}}…
Question Number 159217 by mathlove last updated on 14/Nov/21 Answered by mnjuly1970 last updated on 15/Nov/21 $$\:\:\:\:\:\:\:{solution}.. \\ $$$$\:\:\:\Omega:\overset{{hopital}\:{rule}} {=}{lim}_{\:{x}\rightarrow\frac{\pi}{\mathrm{6}}} \:\frac{\:\mathrm{2}\:.\:\frac{\left(\Gamma\left({sin}\left({x}\right)\right)'\right.}{\Gamma\left({sin}\left({x}\right)\right)}}{\left(\frac{\mathrm{1}}{{sin}\left({x}\right)}\overset{\:\:'} {\right)}\Gamma'\left(\:\frac{\mathrm{1}}{{sin}\left({x}\right)}\right)}\: \\ $$$$\:\:=\mathrm{2}\:{lim}_{\:{x}\rightarrow\frac{\pi}{\mathrm{6}}} \frac{\:{cos}\left({x}\right)\Gamma'\left({sin}\left({x}\right)\right)}{\Gamma\left(\:{sin}\left({x}\right)\right)}\:.\frac{{sin}^{\:\mathrm{2}}…
Question Number 93676 by john santu last updated on 14/May/20 $$\mathrm{solve}\:\mathrm{without}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{5}\sqrt{\mathrm{x}+\mathrm{1}}−\mathrm{2}\sqrt{\mathrm{x}+\mathrm{4}}−\mathrm{1}}{\mathrm{x}}\:? \\ $$ Commented by mathmax by abdo last updated on 14/May/20…
Question Number 93648 by i jagooll last updated on 14/May/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{2x}}{\mathrm{x}^{\mathrm{3}} }\:? \\ $$ Answered by 675480065 last updated on 14/May/20 Answered by john…
Question Number 28098 by tawa tawa last updated on 20/Jan/18 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{log}_{\mathrm{e}} \mathrm{x}}{\mathrm{x}^{\mathrm{h}} }\:,\:\:\:\:\:\:\:\:\:\:\:\mathrm{h}\:>\:\mathrm{0} \\ $$ Answered by ajfour last updated on 20/Jan/18 $$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{1}/{x}\right)}{{hx}^{{h}−\mathrm{1}}…
Question Number 28097 by tawa tawa last updated on 20/Jan/18 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3}^{\mathrm{x}} \:−\:\mathrm{2}^{\mathrm{x}} }{\mathrm{x}^{\mathrm{2}} } \\ $$ Commented by abdo imad last updated on 20/Jan/18…
Question Number 28096 by tawa tawa last updated on 20/Jan/18 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}\:−\:\mathrm{sinx}} \\ $$ Commented by abdo imad last updated on 20/Jan/18 $$=\:{lim}_{{x}\rightarrow\propto\:\:} \:\:\frac{{x}}{\mathrm{1}−\frac{{sinx}}{{x}}}\:\:\:{but}\:\:\:/\frac{{sinx}}{{x}}/\leqslant\:\:\:\frac{\mathrm{1}}{/{x}/}\:_{{x}\rightarrow\propto}…
Question Number 28093 by tawa tawa last updated on 20/Jan/18 $$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\:\left(\mathrm{1}\:+\:\mathrm{tanx}\right)^{\mathrm{cotx}} \\ $$ Commented by abdo imad last updated on 20/Jan/18 $$={lim}_{{x}\rightarrow\mathrm{0}^{−} }…