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${a_0}...{a_k} \ge 0$
$(a_k)$ is a finite sequence.

$$\mathop {\lim }\limits_{n \to \infty } \root n \of {{a_k}{n^k} + {a_{k - 1}}{n^{k - 1}} + ... + {a_1}n + {a_0}} $$

Any help/hint will be appreciated!

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    $\begingroup$ Take $\large\log\mbox{'s}$ and use L'Hopital rule. $\endgroup$ – Felix Marin Nov 14 '13 at 8:59
  • $\begingroup$ I can't use L'Hopital rule. It hasn't been taught yet in our class $\endgroup$ – captain dragon Nov 14 '13 at 9:12
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For the problem to make sense, it is necessary that $a_k> 0$. In these conditions:

$$\mathop {\lim }\limits_{n \to \infty } \root n \of {{a_k}{n^k} + {a_{k - 1}}{n^{k - 1}} + ... + {a_1}n + {a_0}}=$$ $$=\mathop {\lim }\limits_{n \to \infty } \root n \of {{n^k} } \ \root n \of {{a_k} + {a_{k - 1}}{n^{-1}} + ... + {a_1}{n^{-k+1}} + {a_0}{n^{-k}}}=1.1=1$$

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