Question Number 59694 by meme last updated on 13/May/19 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{cos}\left(\sqrt{\left.\mid{x}\mid\right)−\mathrm{1}}\right.}{{x}}=? \\ $$ Commented by maxmathsup by imad last updated on 13/May/19 $${let}\:{A}\left({x}\right)\:=\frac{{cos}\left(\sqrt{\mid{x}\mid}\right)−\mathrm{1}}{{x}}\:\:\:\:{we}\:{have}\:{cos}\left({u}\right)\:\sim\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\left(\:\:{u}\in{V}\left(\mathrm{0}\right)\right)\:\Rightarrow \\…
Question Number 125204 by bemath last updated on 09/Dec/20 Answered by liberty last updated on 09/Dec/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{3}^{\mathrm{4}{x}} \right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}{x}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} =\: \\ $$$$\:\mathrm{9}\:×\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}{x}}…
Question Number 59637 by mathtype last updated on 12/May/19 $$\underset{{x}\rightarrow\infty} {{lim}}\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{{x}} \mid\mathrm{sin}\:{x}\mid \\ $$ Commented by Mr X pcx last updated on 13/May/19 $${there}\:{is}\:{a}\:{problem}\:{here}\:{let}\:{take}\:{x}={n}\pi…
Question Number 190679 by mathlove last updated on 09/Apr/23 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{e}^{{x}} }{{x}^{\mathrm{60}!} }=? \\ $$$${pleas}\:{solve}\:{this} \\ $$ Answered by Frix last updated on 09/Apr/23 $$\frac{\mathrm{e}^{{x}}…
Question Number 59599 by tanmay last updated on 12/May/19 Answered by tanmay last updated on 12/May/19 $${T}_{{r}} =\frac{{tan}\frac{{x}}{\mathrm{2}^{{r}+\mathrm{1}} }\left(\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{r}+\mathrm{1}} }\right)}{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}^{{r}+\mathrm{1}} }} \\ $$$${T}_{{r}}…
Question Number 59542 by Mikael_Marshall last updated on 11/May/19 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\mathrm{ln}\left(\mathrm{cosx}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$${I}\:{have}\:{a}\:{doubt} \\ $$ Commented by kaivan.ahmadi last updated on 11/May/19 $${hop} \\…
Question Number 125056 by benjo_mathlover last updated on 08/Dec/20 $$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{5}\:\sqrt[{\mathrm{3}}]{{x}}\:−\mathrm{3}\:\sqrt[{\mathrm{4}}]{{x}}\:−\sqrt[{\mathrm{6}}]{{x}}\:}{\mathrm{4}\:\sqrt[{\mathrm{3}}]{{x}}\:−\sqrt[{\mathrm{24}}]{{x}}}\:? \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{5}\:\sqrt[{\mathrm{3}}]{{x}}\:−\mathrm{3}\:\sqrt[{\mathrm{4}}]{{x}}\:−\sqrt[{\mathrm{6}}]{{x}}}{\mathrm{4}\:\sqrt[{\mathrm{3}}]{{x}}\:−\:\sqrt[{\mathrm{24}}]{{x}}}\:?\: \\ $$ Answered by bramlexs22 last updated on 08/Dec/20 $$\left(\mathrm{1}\right)\:{let}\:{x}\:=\:{t}^{\mathrm{24}} \\…
Question Number 125049 by bramlexs22 last updated on 08/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}+\mathrm{cos}\:\left(\mathrm{3}{x}\right)−\mathrm{3cosh}\:\left({x}\right)\right)^{\mathrm{4}} }{\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:? \\ $$ Answered by Olaf last updated on 08/Dec/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}+\mathrm{cos3}{x}−\mathrm{3cosh}{x}\right)^{\mathrm{4}} }{\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 124995 by liberty last updated on 07/Dec/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}+\mathrm{sin}\:{x}}\:−\sqrt{\mathrm{1}+\mathrm{cos}\:{x}}}{\mathrm{tan}\:{x}}\:=? \\ $$ Answered by bemath last updated on 07/Dec/20 Answered by Dwaipayan Shikari last…
Question Number 190521 by Rupesh123 last updated on 04/Apr/23 Answered by mehdee42 last updated on 04/Apr/23 $$\mathrm{8}+\mathrm{88}+\mathrm{888}+…+\mathrm{888}…\mathrm{8}=\mathrm{8}\left(\frac{\mathrm{10}−\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{10}^{\mathrm{2}} −\mathrm{1}}{\mathrm{9}}+…+\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{9}}\right) \\ $$$$=\frac{\mathrm{8}}{\mathrm{9}}\left(\mathrm{10}+\mathrm{10}^{\mathrm{2}} +…+\mathrm{10}^{{n}} −{n}\right)=\frac{\mathrm{8}}{\mathrm{9}}\left(\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{10}}{\mathrm{9}}−{n}\right) \\…