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

Question

I started learning infinitesimally math and I have the following question: Is the following sentence true

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

I can see that it tends to $\infty$ but I can't prove it.

How can I prove that this sentence is wrong?

Answer 1

Squeeze theorem. Since $\;n\ge 1\;$ and $\;n^5-2n+7>1\;$ always, we have

$$1\le\sqrt[n]{n^5-2n+7}\le\sqrt[n]{3n^5}=\sqrt[n]3\,\left(\sqrt[n]n\right)^5\xrightarrow[n\to\infty]{}1\cdot1^5=1$$

Answer 2

Consider $$A_n= \sqrt[n]{n^5 -2n + 7} $$ Take logarithms $$\log(A_n)=\frac 1 n \log(n^5-2n+7)=\frac 1 n\Big(\log(n^5)+\log\big(1-\frac {2n-7}{n^5}\big)\Big) $$ Now Taylor for the second logarithm $$\log\big(1-\frac {2n-7}{n^5}\big)=-\frac{2}{n^4}+\frac{7}{n^5}+O\left(\frac{1}{n^7}\right)$$ All of that makes $$\log(A_n)=5\frac{ \log (n)}{n}-\frac{2}{n^5}+\frac{7}{n^6}+O\left(\frac{1}{n^7}\right)$$ So, $\log(A_n)\to 0$ and $A_n\to 1$.

You could even continue writing $$A_n=e^{\log(A_n)}=1+5\frac{ \log (n)}{n}+\frac{25}2\frac{ \log^2 (n)}{n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.