Question Number 188060 by mathlove last updated on 25/Feb/23 Answered by aleks041103 last updated on 26/Feb/23 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\sqrt[{{k}}]{{cos}\left({kx}\right)}\right)^{\mathrm{1}/{x}^{\mathrm{2}} } = \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\underset{{k}=\mathrm{2}}…
Question Number 122462 by benjo_mathlover last updated on 17/Nov/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}}{{x}}\:? \\ $$ Answered by Olaf last updated on 17/Nov/20 $${undeterminate} \\ $$ Commented by…
Question Number 122458 by benjo_mathlover last updated on 17/Nov/20 $$\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{{n}}\:}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}}+\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}}+…+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{n}−\mathrm{1}}+\sqrt{\mathrm{2}{n}+\mathrm{1}}}\:\right)=? \\ $$ Answered by liberty last updated on 17/Nov/20 $$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}}}\:\left(\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2k}−\mathrm{1}}+\sqrt{\mathrm{2k}+\mathrm{1}}}\right)\:= \\…
Question Number 187984 by cortano12 last updated on 24/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 187983 by TUN last updated on 24/Feb/23 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{1}}{{x}+\mathrm{2}}+…+\frac{\mathrm{1}}{\mathrm{2}{x}}\right) \\ $$ Answered by cortano12 last updated on 24/Feb/23 $$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{x}+\mathrm{k}}\right) \\…
Question Number 122338 by benjo_mathlover last updated on 15/Nov/20 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}\sqrt{{x}}\:+\:\mathrm{3}\:\sqrt[{\mathrm{3}\:}]{{x}}\:+\:\mathrm{5}\sqrt[{\mathrm{5}\:}]{{x}}}{\:\sqrt{\mathrm{2}{x}−\mathrm{3}}\:+\:\sqrt[{\mathrm{3}\:}]{\mathrm{3}{x}−\mathrm{2}}}\:?\: \\ $$ Commented by liberty last updated on 16/Nov/20 $$\:\:\:\mathrm{Find}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}\sqrt{\mathrm{x}}\:+\:\mathrm{3}\:\sqrt[{\mathrm{3}}]{\mathrm{x}}\:+\:\mathrm{5}\:\sqrt[{\mathrm{5}}]{\mathrm{x}}}{\:\sqrt{\mathrm{2x}−\mathrm{3}}\:+\:\sqrt[{\mathrm{3}\:}]{\mathrm{3x}−\mathrm{2}}}\:. \\ $$$$\:\:\:\mathrm{Solution}: \\…
Question Number 187859 by TUN last updated on 23/Feb/23 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}−{x}\right) \\ $$ Answered by Rasheed.Sindhi last updated on 23/Feb/23 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}−{x}\right) \\…
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Question Number 56708 by gunawan last updated on 22/Mar/19 $$\mathrm{for}\:\mathrm{all}\:{n}\:\in\:\mathbb{N},\:{f}_{{n}} \left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \\ $$$$\mathrm{for}\:\mathrm{any}\:−\mathrm{1}<{x}<\mathrm{1}.\:\mathrm{If}\:{f}\::\:\left(−\mathrm{1},\mathrm{1}\right)\rightarrow\mathbb{R} \\ $$$$\mathrm{with}\:{f}\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{f}_{{n}\:} \mathrm{of}\:\left(−\mathrm{1},\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {f}\left({x}\right)\:{dx}=… \\…
Question Number 56707 by gunawan last updated on 22/Mar/19 $${f}\left({x}\right)=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\:\:\mathrm{for}\:\mathrm{all}\:{x}\:\in\:\mathbb{R} \\ $$$${x}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{6}}\:\mathrm{for}\:\mathrm{all}\:{n}\:\in\mathbb{N} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{f}\left({x}_{{n}} \right)=… \\ $$ Answered by…
Question Number 56706 by gunawan last updated on 22/Mar/19 $${f}\::\:\left[\mathrm{0},\:\mathrm{1}\right)\:\rightarrow\:\mathbb{R} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}\right)=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)−{f}\left(\frac{{x}}{\mathrm{2}}\right)}{{x}}=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)}{{x}}\:=… \\ $$ Commented by Abdo msup.…