Question Number 128456 by Study last updated on 07/Jan/21 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\left({cosx}\right)^{{logx}} =??? \\ $$ Commented by mr W last updated on 07/Jan/21 $$\rightarrow\mathrm{1}^{−\infty} =\mathrm{1} \\…
Question Number 128439 by Study last updated on 07/Jan/21 $${li}\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{m}}\frac{{cosx}}{\mathrm{1}−{sinx}}=??? \\ $$ Answered by benjo_mathlover last updated on 07/Jan/21 $$\mathrm{let}\:\mathrm{x}=\frac{\pi}{\mathrm{2}}+\mathrm{z}\:\wedge\:\mathrm{z}\rightarrow\mathrm{0} \\ $$$$\:\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\mathrm{z}\right)}{\mathrm{1}−\mathrm{sin}\:\left(\mathrm{z}+\frac{\pi}{\mathrm{2}}\right)}=\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:\mathrm{z}}{\mathrm{1}−\mathrm{cos}\:\mathrm{z}}…
Question Number 128411 by bramlexs22 last updated on 07/Jan/21 $$\:\mathrm{If}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{ln}\:\left({a}+{x}\right)−\mathrm{ln}\:{a}}{{x}}\right)+\:{k}\:\underset{{x}\rightarrow\mathrm{e}} {\mathrm{lim}}\left(\frac{\mathrm{ln}\:{x}−\mathrm{1}}{{x}−{e}}\right)=\mathrm{1} \\ $$$$\mathrm{then}\:{k}\:=\: \\ $$ Answered by Dwaipayan Shikari last updated on 07/Jan/21 $$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 62747 by Mikael last updated on 24/Jun/19 $${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}}…