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Evaluate the limit $\lim\limits_{n \to \infty } \root n \of {{a_k}{n^k} + {a_{k - 1}}{n^{k - 1}} + ... + {a_0}} $
$a_0,...a_k > 0$

Is the following right?

for $n$ sufficiently large:
$$1 = \mathop {\lim }\limits_{n \to \infty } \root n \of {{a_k}{n^k}} \le \mathop {\lim }\limits_{n \to \infty } \root n \of {{a_k}{n^k} + {a_{k - 1}}{n^{k - 1}} + ... + {a_0}} \le \mathop {\lim }\limits_{n \to \infty } \root n \of {k \cdot {a_k}{n^k}} = 1$$

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    $\begingroup$ The general idea is good. On the right, we want to replace $a_k$ by the max of the $a_i$. And there are $k+1$ terms. You should write the inequalities without limits, and then take limits. $\endgroup$ – André Nicolas Mar 22 '14 at 18:31
  • $\begingroup$ Without knowing what the relation between the coefficients $\;a_k\;$ is both inequalities may be pretty wrong... $\endgroup$ – DonAntonio Mar 22 '14 at 18:32
  • $\begingroup$ @DonAntonio, Can you explain? I did mention "for sufficiently large $n$". $\endgroup$ – AndrePoole Mar 22 '14 at 18:36
  • $\begingroup$ $n$ need not be large to set up inequalities, the $a_i$ are positive. $\endgroup$ – André Nicolas Mar 22 '14 at 18:42
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Looks good to me: the key is that $\lim\limits_{n\to\infty} n^{1/n}=1$. You might be more explicit about how large $n$ needs to be.

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