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Question Number 75468 by aliesam last updated on 11/Dec/19

$$\underset{n\to \mathrm{\infty }}{lim}\sqrt{\frac{2n^2-5n+3}{n^5+1}}$$

Commented by MJS last updated on 11/Dec/19

$$P_n\left(x\right)=c_n{x}^n+c_{n-1}{x}^{n-1}...+c_{1}x+c_0=\underset{i=0}{\sum ^n}{c}_i{x}^i$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{P_n\left(x\right)}{P_{n+k}\left(x\right)}=0\mathrm{\forall }n,k\in {\mathbb{N}}^{★}$$$$\Rightarrow$$$$\underset{n\to \mathrm{\infty }}{\lim }\sqrt{\frac{2n^2-5n+3}{n^5+1}}=0$$

Commented by mathmax by abdo last updated on 12/Dec/19

$$wehave{lim}_{n\to +\mathrm{\infty }}\frac{2n^2-5n+3}{n^5+1}={lim}_{n\to +\mathrm{\infty }}\frac{2n^2}{n^5}={lim}_{n\to +\mathrm{\infty }}\frac{2}{n^3}=0\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}\sqrt{\frac{2n^2-5n+3}{n^5+1}}=0$$

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