Question Number 206702 by depressiveshrek last updated on 22/Apr/24
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{cos}{n}+\mathrm{sin}{n}−\mathrm{3}^{{n}} +\mathrm{4}^{{n}} } \\ $$
Answered by Frix last updated on 22/Apr/24
$$−\sqrt{\mathrm{2}}\leqslant\mathrm{cos}\:{n}\:+\mathrm{sin}\:{n}\:\leqslant\sqrt{\mathrm{2}} \\ $$$$\forall{a}\in\mathbb{R}:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}^{{n}} −\mathrm{3}^{{n}} +{a}}\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}^{{n}} −\mathrm{3}^{{n}} }\:=\infty \\ $$
Answered by mathzup last updated on 23/Apr/24
$${u}_{{n}} =\sqrt{{cosn}+{sinn}+\mathrm{4}^{{n}} −\mathrm{3}^{{n}} } \\ $$$$\Rightarrow{u}_{{n}} =\sqrt{\sqrt{\mathrm{2}}{cos}\left({n}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{4}^{{n}} −\mathrm{3}^{{n}} } \\ $$$$=\mathrm{2}^{{n}} \sqrt{\frac{\sqrt{\mathrm{2}}{cos}\left({n}+\frac{\pi}{\mathrm{4}}\right)}{\mathrm{4}^{{n}} }+\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} } \\ $$$$\Rightarrow{u}_{{n}} \sim\mathrm{2}^{{n}} \sqrt{\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} }\sim\mathrm{2}^{{n}} \:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} {u}_{{n}} =+\infty \\ $$