Question Number 91419 by john santu last updated on 30/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{4}} {x}}\left(\mathrm{sin}\:\left(\frac{{x}}{\mathrm{1}+{x}}\right)−\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}\right) \\ $$ Commented by john santu last updated on 01/May/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{4}}…
Question Number 25852 by abdo imad last updated on 15/Dec/17 $${let}\:{s}\:{put}\:{H}_{{n}} \:=\:\mathrm{1}\:+\mathrm{2}^{−\mathrm{1}} +\mathrm{3}^{−\mathrm{1}} +….+{n}^{−\mathrm{1}} \:{and}\:\:\:{U}_{{n}} =\:{H}_{{n}} \:−{ln}\left({n}\right) \\ $$$$\:{prove}\:{that}\:{U}_{{n}} \:{is}\:{convergent}\:{to}\:{a}\:{number}\:\:{s}\:{wish}\:{verify} \\ $$$$\mathrm{0}<{s}<\mathrm{1}\:\:\:\left({s}\:{is}\:{named}\:{number}\:{of}\:{Euler}\:\right) \\ $$ Commented…
Question Number 156891 by cortano last updated on 16/Oct/21 Answered by mindispower last updated on 17/Oct/21 $$\mathrm{1}−{tg}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}^{{k}} }\right)=\frac{{cos}\left(\frac{{x}}{\mathrm{2}^{{k}−\mathrm{1}} }\right)}{{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}^{{k}} }\right)} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}}…
Question Number 91274 by M±th+et+s last updated on 29/Apr/20 $$\underset{{n}\rightarrow\infty} {{lim}n}^{−{n}^{\mathrm{2}} } \left[\left({n}+\mathrm{1}\right)\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)….\left({n}+\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\right)\right]^{{n}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 156788 by mathlove last updated on 15/Oct/21 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…..{x}^{{n}} −{n}}{{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…..{x}^{{m}} −{m}}=?\:\:\:\:\frac{\mathrm{0}}{\mathrm{0}} \\ $$ Answered by puissant last updated on…
Question Number 91182 by jagoll last updated on 28/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}^{\mathrm{6}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3tan}\:^{\mathrm{2}} {x}}{\mathrm{3}{x}^{\mathrm{6}} } \\ $$ Answered by john santu last updated on 28/Apr/20…
Question Number 91158 by jagoll last updated on 28/Apr/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}}\:−\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} } \\ $$ Commented by jagoll last updated on 28/Apr/20 $${thank}\:{you}\:{sir} \\…
Question Number 91157 by jagoll last updated on 28/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3tan}\:\mathrm{4}{x}−\mathrm{4tan}\:\mathrm{3}{x}}{\mathrm{3sin}\:\mathrm{4}{x}−\mathrm{4sin}\:\mathrm{3}{x}}\:=\:? \\ $$ Commented by john santu last updated on 28/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{4}{x}+\frac{\left(\mathrm{4}{x}\right)^{\mathrm{3}} }{\mathrm{3}}\right)−\mathrm{4}\left(\mathrm{3}{x}+\frac{\left(\mathrm{3}{x}\right)^{\mathrm{3}} }{\mathrm{3}}\right)}{\mathrm{3}\left(\mathrm{4}{x}−\frac{\left(\mathrm{4}{x}\right)^{\mathrm{3}}…
Question Number 91154 by jagoll last updated on 28/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{12}−\mathrm{6}{x}^{\mathrm{2}} −\mathrm{12cos}\:{x}}{{x}^{\mathrm{4}} } \\ $$ Commented by jagoll last updated on 28/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{6}\left(\mathrm{2}−{x}^{\mathrm{2}} −\mathrm{2cos}\:{x}\right)}{{x}^{\mathrm{4}}…
Question Number 156683 by cortano last updated on 14/Oct/21 Commented by john_santu last updated on 14/Oct/21 $${it}\:{should}\:{be} \\ $$$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} −\mathrm{1}+\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}−\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}{{x}−\mathrm{1}+\sqrt{{x}+\mathrm{1}}−\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\:. \\…