Question Number 183673 by universe last updated on 28/Dec/22 $$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\:\left[\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2}−\boldsymbol{{a}\mathrm{x}}}\right)\right]^{\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2}−\boldsymbol{\mathrm{bx}}}\right)} \\ $$ Answered by AlexandreT last updated on 28/Dec/22 $${lim}\:\left[{sen}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}−{ax}}\right)−\mathrm{1}\right]{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}−{bx}}\right)…
Question Number 118088 by bobhans last updated on 15/Oct/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}.\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:.\sqrt[{\mathrm{3}\:}]{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{3}} }\:=? \\ $$ Answered by bemath last updated on 15/Oct/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}.\sqrt{\mathrm{x}^{\mathrm{2}}…
Question Number 118040 by mathocean1 last updated on 14/Oct/20 Answered by Ar Brandon last updated on 14/Oct/20 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{2cos}{x}}{\pi−\mathrm{3}{x}}=\frac{\mathrm{0}}{\mathrm{0}}=\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\frac{\mathrm{2sin}{x}}{−\mathrm{3}}=\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\boldsymbol{\div}−\mathrm{3}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$ Answered by bobhans…
Question Number 183528 by universe last updated on 26/Dec/22 $$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\:\mathrm{3}\sqrt{\mathrm{1}+\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{3}} }}\:−\mathrm{1}\right)\:\:=\:\:\:? \\ $$ Commented by TANMAY PANACEA last updated…
Question Number 183483 by cortano1 last updated on 26/Dec/22 $$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\sqrt{{x}}}{\:\sqrt{{x}−\sqrt{{x}−\sqrt{{x}−\sqrt{{x}−…}}}}}\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 117898 by bobhans last updated on 14/Oct/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\:\mathrm{1}}{\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}}\:\left(\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}}\right)−\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}\:\right)\:=? \\ $$ Answered by Lordose last updated on 14/Oct/20 $$\frac{\mathrm{1}}{\mathrm{6}}\:\left(\mathrm{rough}\:\mathrm{work}\right) \\ $$ Terms…
Question Number 183376 by TUN last updated on 25/Dec/22 Commented by CElcedricjunior last updated on 25/Dec/22 $$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\boldsymbol{{nx}}^{\boldsymbol{{n}}} =\boldsymbol{{x}}+\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}^{\mathrm{3}} +…\infty\boldsymbol{{x}}^{\infty} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{{x}}\left(\mathrm{1}+\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} +….+\infty\boldsymbol{{x}}^{\infty−\mathrm{1}}…
Question Number 117820 by islam last updated on 13/Oct/20 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left({x}.\mathrm{sin}\:\frac{\mathrm{1}}{{x}}\right)^{{x}^{\mathrm{2}} } \\ $$ Answered by Dwaipayan Shikari last updated on 13/Oct/20 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left({x}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{6}{x}^{\mathrm{3}} }\right)\right)^{{x}^{\mathrm{2}}…
Question Number 183269 by TUN last updated on 24/Dec/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 183260 by TUN last updated on 24/Dec/22 Answered by greougoury555 last updated on 24/Dec/22 $$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$ Answered by TheSupreme last…