Question Number 57073 by gunawan last updated on 31/Mar/19 $${f}\left({x}\right)={A} \\ $$$${f}\::\:\left[\mathrm{0},\:\infty\right)\: \\ $$$${g}\::\:\left[\mathrm{1},\:\mathrm{0}\right] \\ $$$${g}\left({x}\right)={B} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{{n}} {f}\left({x}\right){g}\left(\frac{{x}}{{n}}\right){dx}=… \\ $$ Commented by…
Question Number 122550 by liberty last updated on 18/Nov/20 $$\:\:\underset{{x}\rightarrow\mathrm{1}/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{x}−\mathrm{1}}{\mathrm{tan}\:\left(\pi\mathrm{x}\right)}\:? \\ $$ Answered by bramlexs22 last updated on 18/Nov/20 $$\:\:\underset{{x}\rightarrow\mathrm{1}/\mathrm{2}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)}{\mathrm{tan}\:\left(\pi{x}\right)}\:= \\ $$$$\:\frac{\mathrm{3}}{\mathrm{2}}×\:\underset{{x}\rightarrow\mathrm{1}/\mathrm{2}}…
Question Number 122548 by bramlexs22 last updated on 18/Nov/20 $$\:\:\underset{{x}\rightarrow\mathrm{1}/\mathrm{2}} {\mathrm{lim}}\:\left(\frac{\mathrm{cot}\:\left(\pi{x}\right)}{\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}}\right)=?\: \\ $$ Answered by liberty last updated on 18/Nov/20 $$\:\underset{{x}\rightarrow\mathrm{1}/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{cot}\:\left(\pi\mathrm{x}\right)}{\left(\mathrm{2x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)}\:=\:\underset{{x}\rightarrow\mathrm{1}/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}.\underset{\left(\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{2}}\right)\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cot}\:\left(\pi\mathrm{x}\right)}{\left(\mathrm{2x}−\mathrm{1}\right)}…
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) \\…