Question Number 186064 by TUN last updated on 31/Jan/23 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left({cos}\mathrm{2}{x}\:−{cos}\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{10}}\right) \\ $$$$ \\ $$ Commented by a.lgnaoui last updated on 31/Jan/23 $$\mathrm{2}{x}={t}\:\:\: \\…
Question Number 186066 by TUN last updated on 31/Jan/23 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\left(\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{7}}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{2}}−{x}+\mathrm{1}\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}−\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} }\right) \\ $$$$ \\ $$ Answered by…
Question Number 54992 by Mikael_Marshall last updated on 15/Feb/19 $$\underset{{n}\rightarrow\infty} {{lim}}\left(\sqrt{{n}^{\mathrm{3}} +{n}^{\mathrm{2}} +{n}+\mathrm{1}}−{n}\right) \\ $$ Answered by kaivan.ahmadi last updated on 15/Feb/19 $$×\frac{\sqrt{{n}^{\mathrm{3}} +{n}^{\mathrm{2}} +{n}+\mathrm{1}}+{n}}{\:\sqrt{{n}^{\mathrm{3}}…
Question Number 120515 by john santu last updated on 01/Nov/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{3}\:\sqrt[{{x}\:}]{\mathrm{2}}\:−\mathrm{2}.\sqrt[{{x}\:}]{\mathrm{3}}\:\right)^{{x}} \:? \\ $$ Answered by bramlexs22 last updated on 01/Nov/20 $$\mathrm{ln}\:\mathrm{L}\:=\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}\:\mathrm{ln}\:\left(\mathrm{3}.\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{x}}} \:−\mathrm{2}.\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{x}}}…
Question Number 120511 by bramlexs22 last updated on 01/Nov/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\left[\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)\right]^{\mathrm{3}} }}{{x}\:\mathrm{sin}\:\left(\sqrt{{x}}\right)}\:?\: \\ $$ Answered by Olaf last updated on 01/Nov/20 $$\frac{\sqrt{\mathrm{arcsin}^{\mathrm{3}} \left(\mathrm{2}{x}\right)}}{{x}\mathrm{sin}\left(\sqrt{{x}}\right)}\:\underset{\mathrm{0}^{+} }…
Question Number 120427 by bramlexs22 last updated on 31/Oct/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{cos}\:{x}+\sqrt{\mathrm{2sin}\:{x}+{x}}}\:−\mathrm{1}}{\mathrm{sin}\:{x}} \\ $$ Answered by john santu last updated on 31/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{cos}\:{x}+\sqrt{\mathrm{2sin}\:{x}+{x}}}−\mathrm{1}}{\mathrm{sin}\:{x}}\:= \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 120425 by bobhans last updated on 31/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}^{{x}} −\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\:? \\ $$ Answered by john santu last updated on 31/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}^{{x}} −\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 120277 by bemath last updated on 30/Oct/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{3}} \:\left\{\sqrt{{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\:−\:{x}\sqrt{\mathrm{2}}\:\right\}\:? \\ $$ Commented by benjo_mathlover last updated on 30/Oct/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{4}}…
Question Number 120268 by bramlexs22 last updated on 30/Oct/20 Answered by mathmax by abdo last updated on 30/Oct/20 $$\mathrm{f}\:\mathrm{continue}\:\mathrm{at}\:\mathrm{x}=\mathrm{0}\:\Rightarrow\mathrm{f}\left(\mathrm{0}\right)=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\mathrm{e}^{\mathrm{1}+\mathrm{x}^{\mathrm{2}} \mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)} \\ $$$$\mathrm{we}\:\mathrm{have}\:−\mathrm{1}\leqslant\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\leqslant\mathrm{1}\:\Rightarrow−\mathrm{x}^{\mathrm{2}} \leqslant\mathrm{x}^{\mathrm{2}}…
Question Number 185776 by TUN last updated on 27/Jan/23 Answered by MJS_new last updated on 27/Jan/23 $${n}!\sim\frac{{n}^{{n}} }{\mathrm{e}^{{n}} }\sqrt{\mathrm{2}\pi{n}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\left(\mathrm{2}{n}\right)!}{{n}!{n}^{{n}} }\right)^{\mathrm{1}/{n}} \:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{4}×\mathrm{2}^{\mathrm{1}/\left(\mathrm{2}{n}\right)}…