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
$$\mathrm{2}{x}={t}\:\:\: \\ $$$$\:\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{10}}\:=\mathrm{cos}{t}−\mathrm{cos}{t}\sqrt{\mathrm{1}+\frac{\mathrm{10}}{{t}^{\mathrm{2}} }} \\ $$$$=\mathrm{cos}{t}\left(\mathrm{1}−\sqrt{\mathrm{1}+\frac{\mathrm{10}}{{t}^{\mathrm{2}} }}\:\right) \\ $$$${x}\rightarrow\infty\:\:\:\:\:\:{t}\rightarrow\infty \\ $$$$\mathrm{1}−\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:}\:\rightarrow\mathrm{0} \\ $$$${donc}\:\:\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \left(\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{10}}\:\right)=\overset{} {\mathrm{0}} \\ $$
Answered by Frix last updated on 01/Feb/23
$$\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{10}}\sim\mathrm{2}{x}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{0} \\ $$