Proving that $\lim \limits_{n\to \infty} \sqrt[n]{n^5-2n+7}=1$

Question

In my assignment I have to prove the following:

$$\lim \limits_{n\to \infty} \sqrt[n]{n^5-2n+7}=1$$

I don't know how to start, I believe it has to do with the squeeze theorem, but I can't be sure.

EDIT: can't use L'Hospital. Thanks,

Alan

Answer 1

Let's assume you know that

$$ \lim\limits_{n\to\infty}\sqrt[n]{n}=1. $$

Then you may write, as $n \to \infty$, $$ \sqrt[n]{7} <\sqrt[n]{n^5-2n+7}<\sqrt[n]{n^5+0+n^5} $$ or $$ \sqrt[n]{7} <\sqrt[n]{n^5-2n+7}<\sqrt[n]{2}\times \left(\sqrt[n]{n}\right)^5 $$ and you may conclude easily.

Answer 2

HINT:

Let $A=\sqrt[n]{n^5-2n+7}$

$\lim_{n\to\infty}\ln A=\lim_{n\to\infty}\dfrac{\ln(n^5-2n+7)}n$ which is of the form $\dfrac00$

hence safely apply L'Hospital

Answer 3

Hint: we have $7\leq n^5-2n+7\leq n^5+7n^5=8n^5$ for big enough $n$. With $$\lim\limits_{n\to\infty}\sqrt[n]{n}=1=\lim\limits_{n\to\infty}\sqrt[n]{p}$$ for $p>0$ look for an application of the sandwich theorem.