Question Number 195618 by cortano12 last updated on 06/Aug/23 $$\:\:\:\:\cancel{\underline{\underbrace{ }}} \\ $$ Answered by MM42 last updated on 06/Aug/23 $$\mathrm{0} \\ $$ Commented by…
Question Number 195569 by York12 last updated on 05/Aug/23 $${a}_{{i}} ,{b}_{{i}} ,{x}_{{i}} {be}\:{reals}\:{for}\:{i}=\mathrm{1},\mathrm{2},\mathrm{3},…,{n},\:{such}\:{that} \\ $$$$\sum_{{i}=\mathrm{1}} ^{{n}} \left[{a}_{{i}} {x}_{{i}} \right]=\mathrm{0}.\:{Prove}\:{that} \\ $$$$\left(\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left[{x}_{{i}} ^{\mathrm{2}} \right]\right)\left(\underset{{i}=\mathrm{1}}…
Question Number 195401 by mnjuly1970 last updated on 01/Aug/23 $$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{calc}\underset{\underset{\Downarrow} {\Downarrow}} {\mathrm{u}late} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{lim}_{\:\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\left(\:\:\mathrm{sin}\left(\mathrm{x}\:\right)\right)^{\:\mathrm{tan}^{\:\mathrm{2}} \left(\mathrm{x}\:\right)} \:\:\:\:=\:?\:\:\:\:\:\:\:\:\begin{array}{|c|c|}\\\\\hline\end{array} \\ $$$$…
Question Number 195395 by cortano12 last updated on 01/Aug/23 Answered by kapoorshah last updated on 01/Aug/23 $${Any}\:{number}\:{power}\:\mathrm{0}\:{is}\:\mathrm{1} \\ $$$$ \\ $$$$\left(\mathrm{1}\:+\:\infty\:+\:\infty\:+\:…\:+\:\infty\right)^{\mathrm{0}} \:=\:\mathrm{1} \\ $$ Terms…
Question Number 195365 by Mr.D.N. last updated on 31/Jul/23 $$\:\: \\ $$$$\:\underset{\mathrm{x}\rightarrow\mathrm{y}} {\overset{\mathrm{lim}} {\:}}\:\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{tany}}{\mathrm{x}−\mathrm{y}} \\ $$$$ \\ $$ Answered by cortano12 last updated on 31/Jul/23…
Question Number 195325 by mathlove last updated on 30/Jul/23 $${prove}\:{that} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{1}}{{x}−\frac{\pi}{\mathrm{2}}}=\mathrm{1} \\ $$ Answered by BaliramKumar last updated on 30/Jul/23 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\frac{\mathrm{d}}{\mathrm{dx}}\left({tan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{1}\right)}{\frac{\mathrm{d}}{\mathrm{dx}}\left({x}−\frac{\pi}{\mathrm{2}}\right)}\:=\:\frac{\mathrm{sec}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\centerdot\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}}…
Question Number 195315 by cortano12 last updated on 30/Jul/23 $$\:\:\:\:\:\:\Subset \\ $$ Answered by horsebrand11 last updated on 30/Jul/23 $$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:^{\mathrm{2}} \mathrm{a}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{a}+\mathrm{tan}\:^{\mathrm{4}} \mathrm{a}\:\mathrm{tan}\:^{\mathrm{2}}…
Question Number 195231 by cortano12 last updated on 28/Jul/23 $$\:\:\:\:\:\begin{array}{|c|}{\:\cancel{\underline{\underbrace{ }}}}\\\hline\end{array} \\ $$ Answered by MM42 last updated on 28/Jul/23 $${lim}_{{x}\rightarrow\mathrm{0}} \:\:\sqrt{\frac{\mathrm{1}−{cos}\sqrt{\pi{x}}}{{x}\left(\mathrm{1}+\sqrt{{cos}\sqrt{\pi{x}}}\right)}}\:\: \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\sqrt{\frac{\frac{\mathrm{1}}{\mathrm{2}}\pi{x}}{{x}\left(\mathrm{1}+\sqrt{\left.{cos}\sqrt{\pi{x}}\right)}\right.}}…
Question Number 195259 by mathlove last updated on 28/Jul/23 $${prove}\:{that} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{e}^{{n}} \centerdot\left({n}!\right)}{{n}^{{n}} \:\sqrt{{n}}}=\sqrt{\mathrm{2}\pi} \\ $$ Commented by Frix last updated on 28/Jul/23 $$\mathrm{Easy}\:\mathrm{because}\:\mathrm{for}\:{n}\rightarrow\infty:\:{n}!\rightarrow\frac{{n}^{{n}}…
Question Number 195195 by Mr.D.N. last updated on 26/Jul/23 $$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{limit}: \\ $$$$\:\underset{\mathrm{x}\rightarrow\mathrm{a}\:} {\overset{\mathrm{lim}} {\:}}\:\:\:\frac{\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{2}}} }\:−\:\frac{\boldsymbol{\mathrm{a}}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\boldsymbol{\mathrm{a}}^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$ Commented by Frix last updated…