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Category: Limits

lim-x-27x-3-3x-2-1-3-8x-3-x-2-1-3-x-3-4x-2-2021-1-3-

Question Number 148600 by EDWIN88 last updated on 29/Jul/21 $$\:\:\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\mathrm{27}{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} }\:+\sqrt[{\mathrm{3}}]{\mathrm{8}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2021}}\:=?\: \\ $$ Answered by bemath last updated on…

lim-x-0-1-1-x-2-cos-x-x-4-

Question Number 82988 by jagoll last updated on 26/Feb/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} \:}\mathrm{cos}\:\mathrm{x}}{\mathrm{x}^{\mathrm{4}} } \\ $$ Commented by jagoll last updated on 26/Feb/20 $$\mathrm{my}\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{3}}.\:\mathrm{that}\:\mathrm{is}\:\mathrm{correct}? \\ $$…

Question-82985

Question Number 82985 by M±th+et£s last updated on 26/Feb/20 Answered by ~blr237~ last updated on 26/Feb/20 $$\:{let}\:{n}\geqslant\mathrm{2}\:,\:\:{b}_{{n}} \neq\mathrm{0}\:\:\:{cause}\:\:{a}+{b}\neq\mathrm{0}\:\:{and}\:\:\mathrm{0}<{a}<{b} \\ $$$$\:{we}\:{have}\:\:\:{a}_{{n}} =\frac{{b}_{{n}} ^{\mathrm{2}} }{{b}_{{n}−\mathrm{1}} }\:\:{and}\:\:{b}_{{n}−\mathrm{1}} =\:\mathrm{2}{a}_{{n}}…

lim-x-0-tan-x-sin-x-x-2-x-

Question Number 82924 by jagoll last updated on 26/Feb/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}−\sqrt{\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}} \\ $$ Commented by jagoll last updated on 26/Feb/20 $$\mathrm{the}\:\mathrm{ans}\::\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Commented…

lim-x-2-x-2-x-6-1-3-x-2-x-2-

Question Number 148439 by liberty last updated on 28/Jul/21 $$\:\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}+\mathrm{2}}\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{6}}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}−\mathrm{2}}\:=? \\ $$ Answered by dumitrel last updated on 28/Jul/21 $$\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{\sqrt{{x}+\mathrm{2}}\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{6}}−\mathrm{2}\right)}{{x}−\mathrm{2}}+\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{\mathrm{2}\left(\sqrt{{x}+\mathrm{2}}−\mathrm{2}\right)}{{x}−\mathrm{2}}−\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{{x}^{\mathrm{2}}…

lim-x-0-1-1-1-x-1-ln-x-

Question Number 82815 by M±th+et£s last updated on 24/Feb/20 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\:\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\frac{\mathrm{1}}{{ln}\left({x}\right)}} }=? \\ $$ Commented by mathmax by abdo last updated on 24/Feb/20 $${let}\:{f}\left({x}\right)=\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{−\frac{\mathrm{1}}{{lnx}}}…

lim-x-0-1-1-x-2-cos-2x-x-2-

Question Number 82800 by jagoll last updated on 24/Feb/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\mathrm{cos}\:\mathrm{2}{x}}{{x}^{\mathrm{2}} } \\ $$ Commented by mathmax by abdo last updated on 24/Feb/20 $${let}\:{f}\left({x}\right)=\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}}…