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lim-n-cosn-sinn-3-n-4-n-




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

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