Question Number 62747 by Mikael last updated on 24/Jun/19 $${Are}\:\boldsymbol{{f}},\:\boldsymbol{{g}}:\:\mathbb{R}\rightarrow\mathbb{R}\:{defined}\:{by} \\ $$$${f}\left({x}\right)=\begin{cases}{\mathrm{0},\:\:\:{x}\:\in\:\mathbb{R}\backslash\mathbb{Q}}\\{{x},\:\:\:{x}\:\in\mathbb{Q}}\end{cases} \\ $$$${g}\left({x}\right)=\begin{cases}{\mathrm{1},\:\:\:{x}=\mathrm{0}}\\{\mathrm{0},\:\:\:\:{x}\neq\mathrm{0}}\end{cases} \\ $$$${show}\:{that}\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{0}\:{and}\:\underset{\boldsymbol{\mathrm{y}}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{y}}\right)=\mathrm{0} \\ $$$${however}\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\right)\:{does}\:{not}\:{exist}. \\ $$ Answered by…
Question Number 62659 by mathmax by abdo last updated on 24/Jun/19 $${calculate}\:\:{lim}_{{n}\rightarrow+\infty} \:\:\:\:\frac{\left({n}!\right)^{{n}} }{{n}^{{n}!} } \\ $$ Commented by abdo mathsup 649 cc last updated…
Question Number 62657 by mathmax by abdo last updated on 24/Jun/19 $${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{ln}\left(\mathrm{1}+{tan}\left(\mathrm{2}{x}\right)\right)−{ln}\left({cos}\left(\mathrm{3}{x}\right)\right)}{{x}^{\mathrm{3}} } \\ $$ Commented by mathmax by abdo last updated on 24/Jun/19…
Question Number 62656 by mathmax by abdo last updated on 24/Jun/19 $${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\:\:\frac{\left({n}+\mathrm{1}\right)^{{n}} }{{n}^{{n}+\mathrm{1}} } \\ $$ Commented by mathmax by abdo last updated on…
Question Number 128185 by liberty last updated on 05/Jan/21 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{x}+\mathrm{3}}{\mathrm{x}−\mathrm{1}}\right)^{\mathrm{x}+\mathrm{3}} \:=?\: \\ $$ Answered by john_santu last updated on 05/Jan/21 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}−\mathrm{1}+\mathrm{4}}{{x}−\mathrm{1}}\right)^{{x}+\mathrm{3}} =\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{4}}{{x}−\mathrm{1}}\right)^{{x}+\mathrm{3}}…
Question Number 128187 by john_santu last updated on 05/Jan/21 $$\:\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}}{\:\sqrt{\mathrm{1}−\mathrm{cos}\:{x}}}\:=?\: \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{e}^{\alpha{x}} −{e}^{\beta{x}} }{\mathrm{sin}\:\alpha{x}−\mathrm{sin}\:\beta{x}}=? \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 62626 by Tawa1 last updated on 23/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{of}\:\:\:\:\:\frac{\mathrm{n}!}{\mathrm{4}^{\mathrm{n}} }\:\:\:\mathrm{as}\:\:\mathrm{n}\:\:\mathrm{approach}\:\mathrm{infinity} \\ $$ Commented by mathmax by abdo last updated on 23/Jun/19 $${let}\:{u}_{{n}} =\frac{{n}!}{\mathrm{4}^{{n}} }\:\:\:\:{we}\:{have}\:{n}!\:\sim{n}^{{n}}…
Question Number 128149 by MathSh last updated on 04/Jan/21 $$\underset{{x}\rightarrow−\mathrm{1}} {{lim}}\left(\frac{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}}{{x}+\mathrm{2}−{x}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{{x}+\mathrm{1}}} =\:? \\ $$ Answered by liberty last updated on 04/Jan/21 $$\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\left(\frac{\mathrm{2x}^{\mathrm{3}}…
Question Number 62571 by Joel122 last updated on 23/Jun/19 $$\mathrm{Find} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left({n}\left(\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{3}\right)^{\mathrm{2}} }\:+\:…\:+\:\frac{\mathrm{1}}{\left(\mathrm{4}{n}−\mathrm{1}\right)^{\mathrm{2}} }\right)\right) \\ $$ Commented by Prithwish sen last updated on…
Question Number 62539 by aliesam last updated on 22/Jun/19 $$\mathrm{li}\underset{\mathrm{x}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{sin}\left(\mathrm{2x}\right)−\mathrm{2sin}\left(\mathrm{x}\right)}{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{3}} \right)−\left(\mathrm{arctan}\left(\mathrm{x}\right)\right)^{\mathrm{3}} } \\ $$ Answered by tanmay last updated on 22/Jun/19 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}}…