Question Number 151111 by alcohol last updated on 18/Aug/21 Answered by Olaf_Thorendsen last updated on 18/Aug/21 $$\mathrm{On}\:\mathrm{peut}\:\mathrm{eventuellement}\:\mathrm{avoir}\:\mathrm{3}\:\mathrm{balles} \\ $$$$\mathrm{noires}\:\mathrm{au}\:\mathrm{cours}\:\mathrm{des}\:\mathrm{3},\:\mathrm{4},\:\mathrm{5},…\:{k}\:\mathrm{premiers} \\ $$$$\mathrm{tirages}\:\mathrm{mais}\:\mathrm{pour}\:\mathrm{en}\:\mathrm{etre}\:\mathrm{certain},\:\mathrm{il}\:\mathrm{faut} \\ $$$$\mathrm{au}\:\mathrm{moins}\:\mathrm{tirer}\:\mathrm{1003}\:\mathrm{balles}. \\ $$$$\mathrm{Au}\:\mathrm{pire},\:\mathrm{on}\:\mathrm{aurait}\:\mathrm{500}\:\mathrm{balles}\:\mathrm{blanches},…
Question Number 85542 by TawaTawa1 last updated on 22/Mar/20 Commented by TawaTawa1 last updated on 22/Mar/20 $$\mathrm{Evaluate}:\:\:\:\:\:\underset{{x}\rightarrow\frac{\mathrm{x}}{\mathrm{2}}} {\mathrm{lim}}\:\:\left(\mathrm{x}\:\:−\:\:\frac{\pi}{\mathrm{2}}\right)\:\mathrm{tan}\:\mathrm{x} \\ $$ Commented by mathmax by abdo…
Question Number 151001 by EDWIN88 last updated on 17/Aug/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} {x}}−\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}}{{x}\:\mathrm{tan}\:{x}}\:=?\: \\ $$$$\:\:\: \\ $$ Answered by john_santu last updated on 17/Aug/21…
Question Number 19920 by lidaye last updated on 18/Aug/17 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\int_{\mathrm{0}} ^{\infty} \mathrm{sin}\:{x}^{{n}} \mathrm{d}{x} \\ $$ Commented by 1kanika# last updated on 18/Aug/17 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{of}\:\mathrm{this}\:\mathrm{question}? \\…
Question Number 85293 by jagoll last updated on 20/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}^{+} \:} {\mathrm{lim}}\:\sqrt{\frac{\mathrm{1}}{\mathrm{x}}}−\sqrt{\frac{\mathrm{1}}{\mathrm{2x}}−\mathrm{2020}}\:? \\ $$ Commented by jagoll last updated on 20/Mar/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{result}?\:\mathrm{0}\:\mathrm{or}\:\infty? \\ $$ Answered…
Question Number 150759 by nimnim last updated on 15/Aug/21 $$\:\:\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 $$\frac{\mathrm{ln}\left({e}+{x}\right)−\mathrm{1}}{{x}}\:=\:\frac{\mathrm{ln}{e}+\mathrm{ln}\left(\mathrm{1}+\frac{{x}}{{e}}\right)−\mathrm{1}}{{x}} \\ $$$$=\:\frac{\mathrm{ln}\left(\mathrm{1}+\frac{{x}}{{e}}\right)}{{x}}\:\underset{\mathrm{0}}…
Question Number 150654 by mnjuly1970 last updated on 14/Aug/21 Answered by Ar Brandon last updated on 14/Aug/21 $${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) \\…
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 $${let}\:{f}\left({x}\right)=\frac{{xtan}\left(\mathrm{2}{x}\right)−\mathrm{2}{x}\:{tan}\left({x}\right)}{\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} }\:\:{we}\:{have}\:…
Question Number 85057 by kushdasbaghar@gmail.com last updated on 18/Mar/20 $${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) \\ $$$${sinx}\:{cos}\left(\frac{\mathrm{1}}{{x}}\right)\sim{xcos}\left(\frac{\mathrm{1}}{{x}}\right)\:{and}\:\mid{xcos}\left(\frac{\mathrm{1}}{{x}}\right)\mid\leqslant\mid{x}\mid\rightarrow\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}}…
Question Number 150567 by DELETED last updated on 13/Aug/21 Answered by DELETED last updated on 13/Aug/21 $$\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}}…