lim-n-2n-2-5n-3-n-5-1-

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

\underset{n \to \infty}{l i m} \sqrt{\frac{2 n^{2} - 5 n + 3}{n^{5} + 1}}

Commented by MJS last updated on 11/Dec/19

P_{n} (x) = c_{n} x^{n} + c_{n - 1} x^{n - 1} \dots + c_{1} x + c_{0} = \underset{i = 0}{\sum^{n}} c_{i} x^{i}

\lim_{x \to \infty} \frac{P_{n} (x)}{P_{n + k} (x)} = 0 \forall n, k \in N^{★}

\Rightarrow

\lim_{n \to \infty} \sqrt{\frac{2 n^{2} - 5 n + 3}{n^{5} + 1}} = 0

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

w e h a v e {l i m}_{n \to + \infty} \frac{2 n^{2} - 5 n + 3}{n^{5} + 1} = {l i m}_{n \to + \infty} \frac{2 n^{2}}{n^{5}} = {l i m}_{n \to + \infty} \frac{2}{n^{3}} = 0 \Rightarrow

{l i m}_{n \to + \infty} \sqrt{\frac{2 n^{2} - 5 n + 3}{n^{5} + 1}} = 0