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Category: Limits

lim-x-0-log-e-x-1-x-please-help-

Question Number 150759 by nimnim last updated on 15/Aug/21 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{log}\left({e}+{x}\right)−\mathrm{1}}{{x}}=? \\ $$$${please}\:{help}.. \\ $$ Answered by Olaf_Thorendsen last updated on 15/Aug/21 $$\frac{\mathrm{ln}\left({e}+{x}\right)−\mathrm{1}}{{x}}\:=\:\frac{\mathrm{ln}{e}+\mathrm{ln}\left(\mathrm{1}+\frac{{x}}{{e}}\right)−\mathrm{1}}{{x}} \\ $$$$=\:\frac{\mathrm{ln}\left(\mathrm{1}+\frac{{x}}{{e}}\right)}{{x}}\:\underset{\mathrm{0}}…

Question-150654

Question Number 150654 by mnjuly1970 last updated on 14/Aug/21 Answered by Ar Brandon last updated on 14/Aug/21 $${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{2}}\right) \\…

lim-x-0-x-tan2x-2x-tan-x-1-cos-2x-2-

Question Number 85061 by M±th+et£s last updated on 18/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}\:{tan}\mathrm{2}{x}−\mathrm{2}{x}\:{tan}\left({x}\right)}{\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} } \\ $$ Commented by mathmax by abdo last updated on 18/Mar/20 $${let}\:{f}\left({x}\right)=\frac{{xtan}\left(\mathrm{2}{x}\right)−\mathrm{2}{x}\:{tan}\left({x}\right)}{\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} }\:\:{we}\:{have}\:…

find-lim-x-0-ln-sin-xcos-1-x-1-if-it-exits-

Question Number 85057 by kushdasbaghar@gmail.com last updated on 18/Mar/20 $${find}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}ln}\:\left(\mathrm{sin}\:{x}\mathrm{cos}\:\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)\:{if}\:{it}\:{exits}. \\ $$ Commented by abdomathmax last updated on 19/Mar/20 $${we}\:{have}\:{ln}\left(\mathrm{1}+{sinx}\:{cos}\left(\frac{\mathrm{1}}{{x}}\right)\right)\sim{sinx}\:{cos}\left(\frac{\mathrm{1}}{{x}}\right)\:\left({x}\rightarrow\mathrm{0}\right) \\ $$$${sinx}\:{cos}\left(\frac{\mathrm{1}}{{x}}\right)\sim{xcos}\left(\frac{\mathrm{1}}{{x}}\right)\:{and}\:\mid{xcos}\left(\frac{\mathrm{1}}{{x}}\right)\mid\leqslant\mid{x}\mid\rightarrow\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}}…

Question-150567

Question Number 150567 by DELETED last updated on 13/Aug/21 Answered by DELETED last updated on 13/Aug/21 $$\left.\mathrm{1}\right).\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3x}^{\mathrm{2}} −\mathrm{4}}{\mathrm{2x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3x}^{\mathrm{2}} /\mathrm{x}^{\mathrm{2}} −\mathrm{4}/\mathrm{x}^{\mathrm{2}} }{\mathrm{2x}^{\mathrm{2}} /\mathrm{x}^{\mathrm{2}}…

lim-x-0-tan-4-x-cot-ln-3-x-1-ln-sin-3-x-cos-2-x-1-sin-x-2-2-2-ln-x-2-1-

Question Number 85021 by M±th+et£s last updated on 18/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{tan}^{\mathrm{4}} \left({x}\right)\:{cot}\left({ln}^{\mathrm{3}} \left({x}+\mathrm{1}\right)\right){ln}\left({sin}^{\mathrm{3}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right)+\mathrm{1}\right)}{{sin}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\:−\sqrt{\mathrm{2}}\right){ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$ Terms of Service Privacy Policy Contact:…

lim-x-0-sin-38x-38sin-x-19x-3-

Question Number 84932 by jagoll last updated on 17/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{38x}−\mathrm{38sin}\:\mathrm{x}}{\mathrm{19x}^{\mathrm{3}} }\:=\: \\ $$ Commented by john santu last updated on 17/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{38cos}\:\mathrm{38x}\:−\:\mathrm{38cos}\:\mathrm{x}}{\mathrm{3}.\mathrm{19x}^{\mathrm{2}} }\:=…

lim-x-x-2-sin-x-x-x-2-1-

Question Number 84873 by john santu last updated on 17/Mar/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} \mathrm{sin}\:\left(\frac{\mathrm{x}!}{\mathrm{x}}\right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}} \\ $$ Commented by john santu last updated on 17/Mar/20 $$\mathrm{yes}.\:\mathrm{i}\:\mathrm{agree}\:\mathrm{sir}…