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Category: Limits

bemath-lim-x-pi-1-2x-2pi-pi-x-cos-2t-dt-1-cos-3t-

Question Number 106869 by bemath last updated on 07/Aug/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:@\mathrm{bemath}@ \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\left[\left(\frac{\mathrm{1}}{\mathrm{2x}−\mathrm{2}\pi}\right)\:\underset{\pi} {\overset{\mathrm{x}} {\int}}\:\frac{\mathrm{cos}\:\mathrm{2t}\:\mathrm{dt}}{\mathrm{1}−\mathrm{cos}\:\mathrm{3t}}\:\right]=? \\ $$ Answered by bobhans last updated on 07/Aug/20 $$\:\:\:\:\:\:\:\:\:\overset{\wedge\mathrm{bobhans}\wedge}…

bobhans-lim-x-0-1-sin-4x-1-3-cos-2x-x-tan-x-

Question Number 106842 by bobhans last updated on 07/Aug/20 $$\:\:\:\:\:\circ\mathrm{bobhans}\circ \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{sin}\:\mathrm{4x}}\:−\mathrm{cos}\:\mathrm{2x}}{\mathrm{x}\:\mathrm{tan}\:\mathrm{x}}\:? \\ $$ Commented by bemath last updated on 07/Aug/20 $$\mathrm{the}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{DNE}\: \\ $$…

Question-172361

Question Number 172361 by Mikenice last updated on 25/Jun/22 Answered by mr W last updated on 26/Jun/22 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{o}\left({x}^{\mathrm{5}} \right)−\left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+{o}\left({x}^{\mathrm{5}} \right)\right)}{{x}^{\mathrm{3}} } \\…

lim-x-0-tanx-sinx-sinx-cos2x-cosx-

Question Number 106810 by Study last updated on 07/Aug/20 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{tanx}−{sinx}}{{sinx}\left({cos}\mathrm{2}{x}−{cosx}\right)}=??? \\ $$ Answered by bemath last updated on 07/Aug/20 $$\:\:\:\:\:@\mathrm{bemath}@ \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}\left(\mathrm{cos}\:\mathrm{2x}−\mathrm{cos}\:\mathrm{x}\right)}\:? \\…

bemath-lim-x-pi-4-x-pi-cos-2-x-pi-pi-2x-cos-x-pi-2-

Question Number 106771 by bemath last updated on 07/Aug/20 $$\:\:\:\:\:\:\:@\mathrm{bemath}@ \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\left[\frac{\mathrm{4}\left(\mathrm{x}−\pi\right)\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}{\pi\left(\pi−\mathrm{2x}\right)\:\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{2}}\right)}\right]=\:? \\ $$ Commented by bemath last updated on 07/Aug/20 Answered by…