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Question Number 206702 by depressiveshrek last updated on 22/Apr/24

$$\underset{n\to \mathrm{\infty }}{\lim }\sqrt{\cos n+\sin n-3^n+4^n}$$

Answered by Frix last updated on 22/Apr/24

$$-\sqrt{2}⩽\cos n+\sin n⩽\sqrt{2}$$$$\mathrm{\forall }a\in \mathbb{R}:\underset{n\to \mathrm{\infty }}{\lim }\sqrt{4^n-3^n+a}=\underset{n\to \mathrm{\infty }}{\lim }\sqrt{4^n-3^n}=\mathrm{\infty }$$

Answered by mathzup last updated on 23/Apr/24

$$u_n=\sqrt{cosn+sinn+4^n-3^n}$$$$\Rightarrow u_n=\sqrt{\sqrt{2}cos(n+\frac{\pi }{4})+4^n-3^n}$$$$=2^n\sqrt{\frac{\sqrt{2}cos(n+\frac{\pi }{4})}{4^n}+1-{\left(\frac{3}{4}\right)}^n}$$$$\Rightarrow u_n\sim 2^n\sqrt{1-{\left(\frac{3}{4}\right)}^n}\sim 2^n\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{u}_n=+\mathrm{\infty }$$

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