Question Number 196399 by tri26112004 last updated on 24/Aug/23 $${a}/\:\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{2}\right)} {\mathrm{lim}}\:\left(\mathrm{1}+{xy}\right)^{\frac{\mathrm{2}}{{x}^{\mathrm{2}} +{xy}}} \\ $$$${b}/\:\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} {\mathrm{lim}}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){sin}\left(\frac{\mathrm{1}}{{xy}}\right) \\ $$$${c}/\underset{\left({x},{y}\right)\rightarrow\left(\infty,\infty\right)} {\mathrm{lim}}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){e}^{−\left({x}+{y}\right)} \\ $$ Answered…
Question Number 196396 by mathlove last updated on 24/Aug/23 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\left(−\mathrm{1}\right)^{{n}} \centerdot{n}\right]=? \\ $$ Commented by mokys last updated on 24/Aug/23 $$\infty \\ $$ Terms…
Question Number 196337 by SaRahAli last updated on 22/Aug/23 Answered by MM42 last updated on 22/Aug/23 $${lim}_{{x}\rightarrow\infty} \:\frac{{ln}\left(\mathrm{1}+\frac{{a}}{{x}}\right)}{\frac{\mathrm{1}}{{x}}}\:\overset{\frac{\mathrm{1}}{{x}}=\:{u}} {=}{lim}_{{u}\rightarrow\mathrm{0}} \:\frac{{ln}\left(\mathrm{1}+{au}\right)}{{u}} \\ $$$${hop}={lim}_{{u}\rightarrow\mathrm{0}} \:\frac{\frac{{a}}{\mathrm{1}+{au}}}{\mathrm{1}}\:={a}\:\checkmark \\ $$…
Question Number 196135 by cortano12 last updated on 18/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196072 by dimentri last updated on 17/Aug/23 $$\:\:\:\Subset \\ $$ Answered by cortano12 last updated on 18/Aug/23 $$\:\:\:\underbrace{\Subset} \\ $$ Answered by jabarsing…
Question Number 195982 by mathlove last updated on 14/Aug/23 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\sqrt{{x}}\right)}−\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{{x}}\right)}\right]=? \\ $$$${with}\:{out}\:{l}'{pital}\:{rule} \\ $$ Answered by MM42 last updated on 14/Aug/23 $${lim}_{{x}\rightarrow\mathrm{1}} \:\left(\frac{\mathrm{1}+\sqrt{{x}}}{\mathrm{2}\left(\mathrm{1}−{x}\right)}−\frac{\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }}{\mathrm{3}\left(\mathrm{1}−{x}\right)}\right)…
Question Number 195947 by cortano12 last updated on 13/Aug/23 Answered by MM42 last updated on 14/Aug/23 $$\frac{\mathrm{2}{cosa}}{{sin}^{\mathrm{3}} {a}} \\ $$ Answered by qaz last updated…
Question Number 195905 by cortano12 last updated on 13/Aug/23 $$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{3}+\mathrm{x}^{\mathrm{4}} }\:−\sqrt{\mathrm{3}+\mathrm{tan}\:^{\mathrm{4}} \mathrm{x}}}{\mathrm{x}^{\mathrm{6}} }\:=? \\ $$ Answered by qaz last updated on 13/Aug/23 $$ \\…
Question Number 195840 by NANIGOPAL last updated on 11/Aug/23 $${find}\:{the}\:{valur} \\ $$$$\:\:{li}\underset{\rightarrow\mathrm{0}} {{m}}\:\frac{{x}−{sinx}}{{x}^{\mathrm{3}} } \\ $$ Commented by kokeb last updated on 11/Aug/23 $${find}\:{the}\:{valur} \\…
Question Number 195772 by dimentri last updated on 10/Aug/23 $$\:\:\:\:\cancel{\underline{\underbrace{ }}}\:\cancel{ } \\ $$ Answered by MM42 last updated on 10/Aug/23 $${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{3}}} \:\:\frac{{cos}\frac{\mathrm{3}{x}}{\mathrm{2}}\:−\mathrm{2}{cos}\frac{\mathrm{3}{x}}{\mathrm{2}}×{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}}{\mathrm{4}{cos}\frac{\mathrm{3}{x}}{\mathrm{2}}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}{cos}\mathrm{3}{x}} \\ $$$$={lim}_{{x}\rightarrow\frac{\pi}{\mathrm{3}}}…