Question Number 175996 by sciencestudent last updated on 10/Sep/22 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{a}^{{x}} −\mathrm{1}}{{x}}=? \\ $$ Answered by Ar Brandon last updated on 10/Sep/22 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{a}^{{x}} −\mathrm{1}}{{x}}=\underset{{x}\rightarrow\mathrm{0}}…
Question Number 175995 by sciencestudent last updated on 10/Sep/22 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{e}^{\mathrm{3}{x}} −\mathrm{1}}{{x}}=? \\ $$ Answered by Ar Brandon last updated on 10/Sep/22 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\mathrm{3}{x}} −\mathrm{1}}{{x}}=\underset{{x}\rightarrow\mathrm{0}}…
Question Number 175994 by sciencestudent last updated on 10/Sep/22 $${li}\underset{{x}\rightarrow\infty} {{m}}\frac{{e}^{{x}} −\mathrm{1}}{{x}}=? \\ $$ Answered by JDamian last updated on 10/Sep/22 $$\infty \\ $$ Answered…
Question Number 110440 by bemath last updated on 29/Aug/20 $$\mathrm{Given}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{f}\left(\mathrm{x}\right)+\frac{\mathrm{1}}{\mathrm{f}\left(\mathrm{x}\right)}\right)\:=\:\mathrm{2}\:,\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right). \\ $$ Answered by john santu last updated on 29/Aug/20 $$\:{let}\:{g}\left({x}\right)\:=\:{f}\left({x}\right)+\frac{\mathrm{1}}{{f}\left({x}\right)}\:\&\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 175967 by infinityaction last updated on 10/Sep/22 $$\:\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{\mathrm{lim}}}\:\left[\boldsymbol{\mathrm{tan}\Psi}+\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\Psi}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\Psi}}{\mathrm{2}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{2}^{\boldsymbol{{n}}} }\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\Psi}}{\mathrm{2}^{\boldsymbol{{n}}} }\right]\:\:\:\:\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 175880 by cortano1 last updated on 08/Sep/22 Answered by blackmamba last updated on 09/Sep/22 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{bx}^{\mathrm{2}} \left(\frac{\mathrm{sin}\:{x}}{{x}}\right)^{\mathrm{2}} +\mathrm{3}{x}\left(\frac{\mathrm{tan}\:\mathrm{3}{x}}{\mathrm{3}{x}}\right)+\mathrm{2}{x}^{\mathrm{2}} \left(\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}{{x}}\right)^{\mathrm{2}} }{\:\sqrt{\mathrm{2}}\:{x}\left(\frac{\mathrm{sin}\:\frac{\mathrm{3}}{\mathrm{2}}{x}}{{x}}\right)+{x}^{\mathrm{2}{b}} \left(\frac{\mathrm{sin}\:\mathrm{3}{x}}{{x}}\right)^{\mathrm{2}{b}} }\: \\…
Question Number 44795 by behi83417@gmail.com last updated on 04/Oct/18 Commented by maxmathsup by imad last updated on 05/Oct/18 $${let}\:{A}\left({x}\right)=\left(\frac{{sinx}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:\Rightarrow{ln}\left({A}_{} \left({x}\right)\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\frac{{sinx}}{{x}}\right)\:{but} \\ $$$${sinx}\:={x}−\frac{{x}^{\mathrm{3}}…
Question Number 110307 by bemath last updated on 28/Aug/20 $$\left(\mathrm{1}\right)\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}−\mathrm{3}{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}}\:? \\ $$$$\left(\mathrm{2}\right)\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\mathrm{arctan}\:\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{1}+{x}−{x}^{\mathrm{2}} }\right)\:{dx} \\ $$$$\left(\mathrm{3}\right){how}\:{many}\:{integer}\:{solution}\:{sets} \\ $$$${exist}\:{for}\:{the}\:{equation}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$…
Question Number 110260 by bemath last updated on 28/Aug/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\:\mathrm{cos}\:\left(\frac{\mathrm{1}}{{x}}\right)\:? \\ $$ Answered by john santu last updated on 28/Aug/20 $${we}\:{know}\:{that}\:−\mathrm{1}\leqslant\mathrm{cos}\:\frac{\mathrm{1}}{{x}}\leqslant\mathrm{1}\: \\ $$$${so}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}\mathrm{cos}\:\left(\frac{\mathrm{1}}{{x}}\right)=\:\infty…
Question Number 175765 by sciencestudent last updated on 06/Sep/22 $${li}\underset{{x}\rightarrow\infty} {{m}}\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)^{−{x}} =? \\ $$ Answered by Ar Brandon last updated on 06/Sep/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)^{−{x}} \\…