Question Number 197660 by pticantor last updated on 25/Sep/23 Answered by witcher3 last updated on 25/Sep/23 $$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}+…\mathrm{t}^{\mathrm{n}} }\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{dt}=\mathrm{1} \\ $$$$\underset{\mathrm{n}\rightarrow\infty}…
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Question Number 197548 by cortano12 last updated on 21/Sep/23 $$\:\:\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{4x}−\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{5}}}{\mathrm{2x}−\mathrm{1}}\right)^{\mathrm{bx}} =?\: \\ $$ Answered by MM42 last updated on 21/Sep/23 $${lim}_{{x}\rightarrow−\infty} \:\left(\frac{\mathrm{4}{x}−\mid\mathrm{2}{x}\mid}{\mathrm{2}{x}−\mathrm{1}}\right)^{{bx}} ={lim}_{{x}\rightarrow−\infty}…
Question Number 197562 by universe last updated on 21/Sep/23 $$\:\:\mathrm{let}\:\:\mathrm{f}_{\mathrm{n}} \left(\mathrm{x}\right)\:=\:\mathrm{nsin}^{\mathrm{2n}+\mathrm{1}} \mathrm{x}\:\mathrm{cos}\:\mathrm{x}\:\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{f}_{\mathrm{n}} \left(\mathrm{x}\right)\:\mathrm{dx}\:−\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left(\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{f}_{\mathrm{n}} \left(\mathrm{x}\right)\right)\mathrm{dx}\:\:\:=\:\:?\: \\ $$ Terms…
Question Number 197514 by cortano12 last updated on 20/Sep/23 $$\:\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sin}\:\mathrm{x}\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)=? \\ $$ Answered by MM42 last updated on 20/Sep/23 $$\frac{\mathrm{1}}{{x}}={t}\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \:{sin}\frac{\mathrm{1}}{{t}}×{sin}^{−\mathrm{1}} {t}\:=\mathrm{0} \\…
Question Number 197483 by cortano12 last updated on 19/Sep/23 $$\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\:\sqrt{\mathrm{1}−\mathrm{x}}}\:=? \\ $$ Answered by Frix last updated on 19/Sep/23 $$=\frac{−\frac{\pi}{\mathrm{2}}}{\:\sqrt{\mathrm{1}+\infty}}=\mathrm{0} \\ $$ Terms…
Question Number 197482 by cortano12 last updated on 19/Sep/23 $$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{3}}\:+\mathrm{2}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{1}}\:\right)=?\: \\ $$ Answered by Frix last updated on 19/Sep/23 $${f}\left({x}\right)=\frac{\sqrt{\mathrm{3}}{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}}…
Question Number 197479 by pticantor last updated on 19/Sep/23 $$\boldsymbol{{find}}: \\ $$$$ \\ $$$$\:\:\:\:\:\:\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{m}}}\:\boldsymbol{{U}}_{\boldsymbol{{n}}} \:=\sqrt[{\mathrm{3}}]{\boldsymbol{{n}}^{\mathrm{3}} +\mathrm{2}\boldsymbol{{n}}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{\boldsymbol{{n}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{n}}^{\mathrm{2}} }\: \\ $$ Commented by Frix…
Question Number 197469 by dragan91 last updated on 18/Sep/23 $$ \\ $$$${solve}\:{limits}\:{for}\:{functions} \\ $$$${f}\left({x}\right)={cos}\left({sgn}\left(\mathrm{1}/{x}\right)\right) \\ $$$${f}\left({x}\right)={sgn}\left({cos}\left(\mathrm{1}/{x}\right)\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ Terms…