Question Number 90077 by manuel__ last updated on 21/Apr/20 $${lim}_{{x}\rightarrow\infty} \left(\mathrm{sin}\:\left({x}+\frac{\mathrm{1}}{{x}}\right)−{sin}\left({x}\right)\right)=? \\ $$ Commented by jagoll last updated on 21/Apr/20 $$\mathrm{sin}\:\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)−\mathrm{sin}\:\left(\mathrm{x}\right)\:=\: \\ $$$$\mathrm{2cos}\:\left(\frac{\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{x}}}{\mathrm{2}}\right)\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2x}}\right)\:= \\ $$$$\mathrm{2cos}\:\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2x}}\right)\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2x}}\right)…
Question Number 90060 by jagoll last updated on 21/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:\mathrm{x}\right)}{\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{2}+\mathrm{x}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\mathrm{3x}}}\:=\:? \\ $$ Commented by john santu last updated on 21/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} +{o}\left({x}^{\mathrm{3}}…
Question Number 24405 by chernoaguero@gmail.com last updated on 17/Nov/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{assymptot} \\ $$$$\underset{{x}\rightarrow\infty\:} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{6}}]{\mathrm{3x}^{\mathrm{2}} +\mathrm{4}}}{\:\sqrt[{\mathrm{9}}]{\mathrm{1}−\mathrm{2x}^{\mathrm{3}} }} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 24404 by moxhix last updated on 17/Nov/17 $${put}\:{a}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}{n}}\centerdot\frac{\mathrm{3}}{\mathrm{2}{n}}\centerdot\frac{\mathrm{5}}{\mathrm{2}{n}}\centerdot\:…\:\centerdot\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}\centerdot{e}^{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {{lim}a}_{{n}} =\:? \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 89922 by student work last updated on 20/Apr/20 Commented by john santu last updated on 20/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt{\mathrm{5}−\mathrm{2}{x}^{\mathrm{2}} }\:\leqslant\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}\right)\:\leqslant\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\sqrt{\mathrm{5}+\mathrm{2}{x}^{\mathrm{2}} } \\…
Question Number 89882 by student work last updated on 19/Apr/20 Answered by Joel578 last updated on 19/Apr/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}!\left(\mathrm{1}−\:{x}\:−\:\mathrm{1}\right)}{\left(\mathrm{2}{x}−\mathrm{1}\right){x}!}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{−{x}}{\mathrm{2}{x}−\mathrm{1}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by TANMAY…
Question Number 155392 by cortano last updated on 30/Sep/21 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\left({a}^{{x}+\mathrm{1}} +{b}^{{x}+\mathrm{1}} \right)^{\mathrm{2}} }{{a}+{b}}\right)^{\frac{\mathrm{1}}{{x}}} =\ldots? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 155357 by mathlove last updated on 29/Sep/21 Commented by mathlove last updated on 30/Sep/21 $${pleas}\:{answer} \\ $$ Answered by qaz last updated on…
Question Number 89712 by thor123 last updated on 18/Apr/20 Commented by mathmax by abdo last updated on 18/Apr/20 $${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }{{x}−\mathrm{2}}\:=\frac{−\mathrm{2}^{\mathrm{2}} }{−\mathrm{2}}\:=\mathrm{2} \\ $$…
Question Number 155191 by mathlove last updated on 26/Sep/21 Terms of Service Privacy Policy Contact: info@tinkutara.com