Question Number 205307 by universe last updated on 15/Mar/24 $$\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\lfloor{a}\rfloor+\lfloor\mathrm{2}{a}\rfloor+…+\lfloor{na}\rfloor}{{n}^{\mathrm{2}} }\:\mathrm{where}\:{a}\in\mathbb{R} \\ $$$$\:\:\:\mathrm{and}\:\lfloor{x}\rfloor\:\mathrm{is}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{of}\:\mathrm{x}\:\in\:\mathbb{R} \\ $$ Commented by Frix last updated on 15/Mar/24 $$\mathrm{Just}\:\mathrm{guessing}: \\…
Question Number 205237 by universe last updated on 13/Mar/24 Answered by Berbere last updated on 13/Mar/24 $${n}^{\mathrm{2}} +{x}^{\mathrm{2}} \geqslant{n}^{\mathrm{2}} \\ $$$$\frac{{x}}{\mathrm{1}+{x}}\leqslant\mathrm{1}\Rightarrow\frac{{nx}\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{\left(\mathrm{1}+{x}\right)\left({n}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)}\leqslant{n}.\mathrm{1}.\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{n}^{\mathrm{2}}…
Question Number 205142 by universe last updated on 10/Mar/24 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{n}^{−\mathrm{n}^{\mathrm{2}} } \left[\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)…\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }\right)\right]^{\mathrm{n}} =? \\ $$ Answered by pi314 last updated on 10/Mar/24…
Question Number 205134 by universe last updated on 09/Mar/24 Answered by pi314 last updated on 09/Mar/24 $${nx}={y} \\ $$$$\Leftrightarrow{A}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{{n}} \frac{{f}\left(\frac{{y}}{{n}}\right)}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}{dy}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{{n}}…
Question Number 205114 by 2kdw last updated on 09/Mar/24 $${Solve}: \\ $$$$ \\ $$$$\:\:{lim}_{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} \frac{\mathrm{1}−{cos}\left(\sqrt{\mathrm{10}{xy}}\right)}{\mathrm{3}.{y}.{sin}\left(\mathrm{22}{x}\right)} \\ $$$$ \\ $$$${Ans}.:\:\frac{\mathrm{5}}{\mathrm{66}} \\ $$$${Step}\:{by}\:{step},\:{please}! \\ $$ Answered by…
Question Number 204991 by tigrecomplexe last updated on 04/Mar/24 Commented by tigrecomplexe last updated on 05/Mar/24 $${sommeone}\:{can}\:{put}\:{hand}\:{here}\:{please}\:? \\ $$ Answered by witcher3 last updated on…
Question Number 204929 by universe last updated on 02/Mar/24 $$\:\:\:\:\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\prod}}\:\frac{{n}^{\mathrm{2}} −{r}}{{n}^{\mathrm{2}} +{r}}\:\:=\:\:? \\ $$ Answered by witcher3 last updated on 02/Mar/24 $$\underset{\mathrm{r}=\mathrm{1}}…
Question Number 204905 by Manishkumar last updated on 02/Mar/24 $$\mathrm{4}.\:\frac{\mathrm{sin}\:\mathrm{30}^{°} \:+\:\mathrm{tan}\:\mathrm{45}^{°} \:−\:\mathrm{cosec}\:\mathrm{60}^{°} }{\mathrm{sec}\:\mathrm{30}^{°} \:+\:\mathrm{cos}\:\mathrm{60}^{°} \:+\:\mathrm{cot}\:\mathrm{45}^{°} } \\ $$$$ \\ $$$$=\:\frac{\mathrm{1}/\mathrm{2}\:+\:\mathrm{1}\:−\:\mathrm{2}/\sqrt{\mathrm{3}}}{\mathrm{2}/\sqrt{\mathrm{3}}\:+\:\mathrm{1}/\mathrm{2}\:+\:\mathrm{1}\:\:\mathrm{v}} \\ $$$$=\:\frac{\frac{\sqrt{\mathrm{3}}\:+\:\mathrm{2}\sqrt{\mathrm{3}}\:−\:\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{3}}}}{\frac{\mathrm{4}\:+\:\sqrt{\mathrm{3}}\:+\:\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{3}}}} \\ $$$$ \\…
Question Number 204900 by universe last updated on 01/Mar/24 $$\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{n}!\left({e}−\mathrm{x}_{\mathrm{n}} \right)\:=\:? \\ $$$$\:\:\mathrm{where}\:\mathrm{x}_{\mathrm{n}\:} =\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}+…+\frac{\mathrm{1}}{\mathrm{n}!} \\ $$ Commented by Frix last updated on 01/Mar/24 $${x}_{{n}}…
Question Number 204879 by universe last updated on 09/Aug/24 $$\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{n}!\left({e}−\mathrm{x}_{\mathrm{n}} \right)\:=\:? \\ $$$$\:\:\mathrm{where}\:\mathrm{x}_{\mathrm{n}\:} =\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}+…+\frac{\mathrm{1}}{\mathrm{n}!} \\ $$ Commented by mr W last updated on 01/Mar/24…