Question Number 179271 by cortano1 last updated on 27/Oct/22 $$\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\sqrt[{\mathrm{3}}]{\mathrm{n}^{\mathrm{3}} +\frac{\mathrm{n}}{\mathrm{3}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{n}}\right)\:\right]^{\mathrm{n}^{\mathrm{3}} } =? \\ $$$$\:\: \\ $$ Answered by mr W last updated on…
Question Number 113651 by bemath last updated on 14/Sep/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}+{e}^{−{x}} }\right).\frac{\mathrm{1}}{\mathrm{3}{x}}\:=\:? \\ $$ Answered by john santu last updated on 14/Sep/20 $${by}\:{Taylor}\:{series}\: \\ $$$${let}\:{f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+{e}^{−{x}}…
Question Number 113601 by bobhans last updated on 14/Sep/20 Answered by bemath last updated on 14/Sep/20 Commented by bobhans last updated on 14/Sep/20 $$\mathrm{santuyy} \\…
Question Number 113598 by bobhans last updated on 14/Sep/20 Answered by bemath last updated on 14/Sep/20 $${let}\:{me}\:{solve} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{3}\:}]{\mathrm{27}{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}} }\right)}−\sqrt[{\mathrm{4}\:}]{\mathrm{64}{x}^{\mathrm{4}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{3}} }\right)}\:= \\…
Question Number 113576 by bemath last updated on 14/Sep/20 $$\:\:\:\underset{{r}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{r}^{\mathrm{2}} +\mathrm{2}{r}}\:−\sqrt[{\mathrm{3}\:}]{\mathrm{8}{r}^{\mathrm{3}} +\mathrm{4}{r}^{\mathrm{2}} }\:=? \\ $$ Answered by bemath last updated on 14/Sep/20 $$\Leftrightarrow\:\underset{{r}\rightarrow\infty} {\mathrm{lim}}{r}\left(\sqrt{\mathrm{4}+\frac{\mathrm{2}}{{r}}}−\sqrt[{\mathrm{3}\:}]{\mathrm{8}+\frac{\mathrm{4}}{{r}}}\right)\:=…
Question Number 47960 by Meritguide1234 last updated on 17/Nov/18 Answered by ajfour last updated on 17/Nov/18 $${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{i}+{j}}{{i}^{\mathrm{2}} +{j}^{\mathrm{2}} } \\…
Question Number 113438 by bemath last updated on 13/Sep/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\left(\mathrm{5}^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}\right)\:? \\ $$ Answered by bemath last updated on 13/Sep/20 Answered by JDamian last…
Question Number 178942 by cortano1 last updated on 23/Oct/22 Answered by mr W last updated on 23/Oct/22 Commented by mr W last updated on 23/Oct/22…
Question Number 113274 by bemath last updated on 12/Sep/20 $$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}+\mathrm{4tan}\:\mathrm{2x}−\mathrm{3tan}\:\mathrm{3x}}{\mathrm{x}^{\mathrm{2}} \:\mathrm{tan}\:\mathrm{x}} \\ $$$$\:\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}}−\sqrt{\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }\: \\ $$$$\:\left(\mathrm{3}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}}+\sqrt{\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\frac{\mathrm{5}}{\mathrm{2}}} } \\ $$ Commented by bemath…
Question Number 178751 by mathlove last updated on 21/Oct/22 Answered by cortano1 last updated on 21/Oct/22 $$\:\mathrm{With}\:\mathrm{L}'\mathrm{Hopital}\:\mathrm{rule} \\ $$$$\:\mathrm{L}=\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{2020x}^{\mathrm{2019}} −\mathrm{2021}}{\mathrm{3x}^{\mathrm{2}} } \\ $$$$\:=\frac{−\mathrm{2020}−\mathrm{2021}}{\mathrm{3}}=\:−\frac{\mathrm{4041}}{\mathrm{3}}=−\mathrm{1347} \\…