Question Number 131383 by EDWIN88 last updated on 04/Feb/21 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}^{\mathrm{2}} }{{x}−\mathrm{1}}\right)^{\mathrm{tan}\:\left(\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)} =? \\ $$ Answered by liberty last updated on 04/Feb/21 $$\mathrm{L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}−\mathrm{1}}\right)^{\mathrm{tan}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\right)}…
Question Number 131373 by EDWIN88 last updated on 10/Feb/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} }\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}}\:−\mathrm{2}{x}\:\right]=? \\ $$ Answered by aleks041103 last updated on 04/Feb/21 $${L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\left[\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}}…
Question Number 131369 by liberty last updated on 04/Feb/21 $$\mathrm{If}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} }\:=\:\frac{\mathrm{m}}{\mathrm{n}}\:\mathrm{where}\:\mathrm{m}\:\mathrm{and}\:\mathrm{n} \\ $$$$\mathrm{are}\:\mathrm{relative}\:\mathrm{prime}\:\mathrm{positive}\:\mathrm{integer}\: \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\left(\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} \right)\:\mathrm{equals} \\ $$ Answered by EDWIN88 last updated…
Question Number 131371 by EDWIN88 last updated on 04/Feb/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{tan}\:{x}−\mathrm{sin}\:{x}\right)−{x}^{\mathrm{3}} }{{x}^{\mathrm{5}} }\:=? \\ $$ Answered by liberty last updated on 04/Feb/21 $$\mathrm{let}\:\mathrm{x}\:=\:\mathrm{2t}\: \\ $$$$\mathrm{L}=\underset{\mathrm{t}\rightarrow\mathrm{0}}…
Question Number 131370 by EDWIN88 last updated on 10/Feb/21 $$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{2}} }\:+{x}\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}}\:=? \\ $$ Answered by JDamian last updated on 04/Feb/21 $$\infty \\…
Question Number 280 by arnav last updated on 25/Jan/15 $$\mathrm{Evaluate}\:\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{\left(\pi/\mathrm{4}−{x}\right)\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)} \\ $$ Answered by 123456 last updated on 18/Dec/14 $$\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{\left(\frac{\pi}{\mathrm{4}}−{x}\right)\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)}\rightarrow\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$=\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\frac{−\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}{−\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)+\left(\frac{\pi}{\mathrm{4}}−{x}\right)\left(−\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)}…
Question Number 131328 by rs4089 last updated on 03/Feb/21 Commented by MJS_new last updated on 04/Feb/21 $$\mathrm{really}? \\ $$$$\mathrm{just}\:\mathrm{try}\:\mathrm{some}\:\mathrm{values}\:\mathrm{with}\:{n}={k}^{\mathrm{2}} \wedge{k}\in\mathbb{N} \\ $$$${y}_{{k}} =\frac{\left(\mathrm{2}{k}^{\mathrm{2}} \right)!}{\mathrm{2}^{{k}} }…
Question Number 222 by 123456 last updated on 25/Jan/15 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}}{\mathrm{arcsin}\:{x}} \\ $$ Answered by ghosea last updated on 16/Dec/14 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}}{\mathrm{arcsin}\:{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}}{{x}}\centerdot\frac{{x}}{\mathrm{arcsin}\:{x}}…
Question Number 131246 by fajar123_ last updated on 03/Feb/21 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\mathrm{2}{x}+\mathrm{3}\:=\mathrm{2}.\mathrm{2}+\mathrm{3}=\mathrm{4}+\mathrm{3}=\mathrm{7} \\ $$$${jadi},\:{nilai}\:{dari}\:{limit}\:{tersebut} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ Terms of Service Privacy…
Question Number 65700 by Masumsiddiqui399@gmail.com last updated on 02/Aug/19 Commented by Prithwish sen last updated on 02/Aug/19 $$=\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\left[\mathrm{1}+\left(\frac{\mathrm{k}}{\mathrm{n}}\right)^{\mathrm{2}} \right]}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{tan}^{−\mathrm{1}}…