Question Number 128567 by mnjuly1970 last updated on 08/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:\:{calculus}… \\ $$$$\:{prove}\:{that}:::: \\ $$$$\:\:\:\phi=\int_{\mathrm{0}} ^{\:\infty} {sin}\left({x}^{\mathrm{2}} \right){sin}\left({x}^{−\mathrm{2}} \right){dx} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\frac{\pi}{\mathrm{2}}}\:\left({e}^{−\mathrm{2}} +{sin}\left(\mathrm{2}\right)−{cos}\left(\mathrm{2}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$…
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Question Number 128456 by Study last updated on 07/Jan/21 $${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}}…