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Question Number 186764 Answers: 0 Comments: 3
Question Number 186762 Answers: 3 Comments: 4
Question Number 186758 Answers: 2 Comments: 0
$$\mathrm{If}\:\left[{t}\right]\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{part}\:\mathrm{of}\:{t},\:\mathrm{then}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[{x}\:\mathrm{sin}\:\pi{x}\right] \\ $$$$\left(\mathrm{A}\right)\:\:\mathrm{equals}\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\:\mathrm{equals}\:−\mathrm{1} \\ $$$$\left(\mathrm{C}\right)\:\:\mathrm{equals}\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{does}\:\mathrm{not}\:\mathrm{exist} \\ $$
Question Number 186752 Answers: 1 Comments: 0
$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{7}}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{4}} +\mathrm{3}}}{\:\sqrt[{\mathrm{4}}]{{x}^{\mathrm{6}} +\mathrm{2}{x}^{\mathrm{5}} +\mathrm{1}}−\sqrt[{\mathrm{5}}]{{x}^{\mathrm{7}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}}} \\ $$$${Please}\:{show}\:{work}. \\ $$
Question Number 186751 Answers: 3 Comments: 0
Question Number 186750 Answers: 1 Comments: 0
Question Number 186748 Answers: 0 Comments: 0
$$\mathrm{Let}\:{f}:\mathbb{R}^{+} \rightarrow\mathbb{R}^{+} \:\mathrm{be}\:\mathrm{a}\:\mathrm{function}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{relation} \\ $$$${f}\left({x}.{f}\left(\mathrm{y}\right)\right)={f}\left({x}\mathrm{y}\right)+{x}\:\mathrm{for}\:\mathrm{all}\:{x},\:\mathrm{y}\:\in\mathbb{R}^{+} .\:\mathrm{Then} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\left({f}\left({x}\right)\right)^{\mathrm{1}/\mathrm{3}} −\mathrm{1}}{\left({f}\left({x}\right)\right)^{\mathrm{1}/\mathrm{2}} −\mathrm{1}}\right)= \\ $$$$\left(\mathrm{A}\right)\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{C}\right)\:\:\frac{\mathrm{2}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Question Number 186741 Answers: 1 Comments: 0
$$\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{18}}\right).\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{18}}\right).\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{18}}\right).\mathrm{cos}\:\left(\frac{\mathrm{7}\pi}{\mathrm{18}}\right)=? \\ $$
Question Number 186739 Answers: 1 Comments: 0
$$\sqrt[{\mathrm{3}}]{\mathrm{5}+\sqrt[{\mathrm{3}}]{\mathrm{5}+\sqrt[{\mathrm{3}}]{\mathrm{5}+...}}}=? \\ $$
Question Number 186737 Answers: 1 Comments: 1
Question Number 186736 Answers: 1 Comments: 0
Question Number 186735 Answers: 1 Comments: 0
Question Number 186726 Answers: 2 Comments: 0
Question Number 186721 Answers: 1 Comments: 1
Question Number 186705 Answers: 1 Comments: 1
$$\mathrm{a},\mathrm{b}>\mathrm{0}\:,\:\mathrm{a}+\mathrm{b}=\mathrm{2} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{a}^{\mathrm{2b}} +\mathrm{b}^{\mathrm{2a}} +\left(\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\mathrm{2} \\ $$
Question Number 186701 Answers: 2 Comments: 2
Question Number 186698 Answers: 0 Comments: 0
Question Number 186697 Answers: 2 Comments: 0
Question Number 186692 Answers: 1 Comments: 1
Question Number 186691 Answers: 1 Comments: 0
Question Number 186690 Answers: 1 Comments: 0
Question Number 186689 Answers: 1 Comments: 1
$$\underset{{n}={o}} {\overset{+{oo}} {\sum}}\:\frac{{x}^{\mathrm{3}{n}} }{\left(\mathrm{2}{n}\right)!}\:\:=\:\:\:? \\ $$
Question Number 186688 Answers: 0 Comments: 1
$${a}_{\mathrm{1}} =\mathrm{0} \\ $$$${a}_{\mathrm{2}} =\mathrm{1} \\ $$$${a}_{{n}+\mathrm{2}} ={a}_{{n}+\mathrm{1}} −{a}_{{n}} \\ $$$${a}_{\mathrm{885}} =? \\ $$
Question Number 186685 Answers: 1 Comments: 0
$$\left(\mathrm{sin}{x}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}{x}−\mathrm{2}\left(\mathrm{cos}{x}\right)^{\mathrm{2}} \geq\mathrm{0} \\ $$$${x}\in\left[\mathrm{0};\mathrm{2}\pi\right] \\ $$
Question Number 186682 Answers: 0 Comments: 0
$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{set}\:\mathrm{identities} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{A}\cup\left(\mathrm{B}\cup\mathrm{C}\right)=\left(\mathrm{A}\cup\mathrm{B}\right)\cup\mathrm{C} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{A}\cup\varnothing=\mathrm{A} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{A}\cap\left(\mathrm{B}\cup\mathrm{C}\right)=\left(\mathrm{A}\cap\mathrm{B}\right)\cup\left(\mathrm{A}\cap\mathrm{C}\right) \\ $$
Question Number 186681 Answers: 0 Comments: 0
$${Etudier}\:{la}\:{convergence}\:{uniforme} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\left(−\right)^{{n}} \frac{{e}^{−{nx}^{\mathrm{2}} } }{\left(\mathrm{1}+{n}\right)^{\mathrm{3}} }\:;\:{n}\:\in\:\mathbb{N}. \\ $$
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