Evaluate $\lim_{n\to\infty}({n^5-2n+7})^{\frac{1}{n}}$

Question

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?

Answer 1

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.

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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 $n \to \infty$.

Answer 2

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.

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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, $$\lim_{n \to \infty} {n^{1/n}} = 1$$

Answer 3

Here are the steps $$ \lim_{n\to\infty}({n^5-2n+7})^{\frac{1}{n}}=\lim_{n\to\infty} e^{\ln(n^5-2n+7)^{\frac{1}{n}}}= \lim_{n\to\infty} e^{\large\frac{\ln(n^5-2n+7)}{n}}= e^{\lim\limits_{n\to\infty}\large\left[\frac{\ln(n^5-2n+7)}{n}\right]}=e^{\lim\limits_{n\to\infty}\large\left[\frac{\frac{d}{dn}\ln(n^5-2n+7)}{\frac{d}{dn}n}\right]} = e^{\lim\limits_{n\to\infty}\large\left[\frac{\frac{1}{(n^5-2n+7)}\frac{d}{dn}[n^5-2n+7]}{1}\right]} = e^{\lim\limits_{n\to\infty}\large\left[\frac{5n^4-2}{n^5-2n+7}\right]} = e^{\lim\limits_{n\to\infty}\large\left[\frac{\frac{5}{n}-\frac{2}{n^5}}{1-\frac{2}{n^4}+\frac{7}{n^5}}\right]}=e^{\large\left[\frac{0-0}{1-0+0}\right]}=e^0=1 $$