Evaluate $\lim_{n\to\infty}({n^5-2n+7})^{\frac{1}{n}}$ I have the following series:

$$ a_n= ({n^5-2n+7})^{\large\frac{1}{n}}$$

I need to evaluate
$$\lim_{n\to\infty}a_n=\lim_{n\to\infty}({n^5-2n+7})^{\large\frac{1}{n}} $$
What's the trick here?
 A: For $n>4$, we have $1 < (n^5-2n+7)^{1/n} < n^{5/n}$. We have $\lim_{n \to \infty} n^{1/n} = 1$. Now use squeeze theorem to conclude the limit.

Showing $n^{1/n} \to 1$ without L'Hospital rule. We have $n^{1/n}>1$ for all $n \in \mathbb{Z}^+$. Hence, $n^{1/n} = 1+\epsilon_n$ for some $\epsilon_n > 0$.
$$n^{1/n} = 1+\epsilon_n \implies n = \underbrace{(1+\epsilon_n)^n > 1 + \dfrac{n(n-1)}2\epsilon_n^2}_{\text{Binomial theorem}}$$
This gives us
$$(n-1) > \dfrac{n(n-1)}2\epsilon_n^2 \implies \epsilon_n < \sqrt{\dfrac2n}$$
Hence,
$$1 < n^{1/n} = 1+\epsilon_n < 1 + \sqrt{\dfrac2{n}}$$
Hence,
$$\lim_{n \to \infty} {n^{1/n}} = 1$$
A: This limit falls into the category of indeterminant forms. In particular the limit looks like $\infty^0$ from which we cannot deduce an answer.
What we do instead is find the limit of the logarithm of the sequence: $$\ln(a_n) = \frac{\ln(n^5-2n+7)}{n}$$ which gives us the indeterminant form $\infty/\infty$. Apply l'Hopital's rule to find $$\lim \ln(a_n) = \lim \frac{\ln(n^5-2n+7)}{n} = \lim \frac{ \frac{5n^4-2}{n^5-2n+7}}{1} = \frac01$$
This last step holds because the degree of the leading term on the bottom of the upper fraction is larger than the numerator.

To avoid using l'Hopital's rule, we can try the following:
$$\lim \ln(a_n) = \lim \frac{\ln(n^5-2n+7)}{n} = \lim \frac{\ln(n^5(1-2n^{-4}+7n^{-5})}{n} = \lim \left( \frac{5\ln(n)}{n} + \frac{\ln(1-2n^{-4}+7n^{-5})}{n}\right)$$
The second fraction goes to zero as $n \to \infty$. Thus we are left with finding the limit of $ln(n)/n$ as $n\to\infty$.
This is the same as the limit: $\lim_{x\to\infty} \frac{ln(x)}{x}$. For convenience call $f(x) = \ln(x)/x$. If we take the derivative of $f$ we see that $f'(x) = (1-\ln(x))/x^2$ which is negative when $x \ge e$. Thus $f$ is a decreasing function for large enough $x$. It is also a positive function, which means it is bounded below by zero.
Now we if we can find a collection of integers that tend to infinity, and show that $f(n_k) \to 0$ for that sequence we win.
Recall that $\ln(n) = \log_{10}(n)/\log_{10}(e)$. If we consider the integers, $10^k$ we find $$f(10^k) = \frac{1}{\log_{10}(e)} \frac{\log_{10}(10^k)}{10^k} = \frac{1}{\log_{10}(e)} \frac{k}{10^k}.$$ This last term goes to zero as $k \to \infty$, since exponetial functions grow much faster than linear functions.
Thus we can conclude that $\ln(n)/n \to 0$ as $n \to \infty$.
A: Here are the steps
$$ \lim_{n\to\infty}({n^5-2n+7})^{\frac{1}{n}}=\lim_{n\to\infty} e^{\ln(n^5-2n+7)^{\frac{1}{n}}}= \lim_{n\to\infty} e^{\large\frac{\ln(n^5-2n+7)}{n}}= e^{\lim\limits_{n\to\infty}\large\left[\frac{\ln(n^5-2n+7)}{n}\right]}=e^{\lim\limits_{n\to\infty}\large\left[\frac{\frac{d}{dn}\ln(n^5-2n+7)}{\frac{d}{dn}n}\right]} = e^{\lim\limits_{n\to\infty}\large\left[\frac{\frac{1}{(n^5-2n+7)}\frac{d}{dn}[n^5-2n+7]}{1}\right]} = e^{\lim\limits_{n\to\infty}\large\left[\frac{5n^4-2}{n^5-2n+7}\right]} = e^{\lim\limits_{n\to\infty}\large\left[\frac{\frac{5}{n}-\frac{2}{n^5}}{1-\frac{2}{n^4}+\frac{7}{n^5}}\right]}=e^{\large\left[\frac{0-0}{1-0+0}\right]}=e^0=1 $$
