0
$\begingroup$

Find $$\lim_{n \to \infty} \frac{(\sqrt[n]{(7^n+n)}-\frac{1}{7})^n}{7^n-n^7}$$

I know that it can be done with using the squeeze theorem but I cannot find a proper upper bound limit

$\endgroup$
3
$\begingroup$

You may write $$ \begin{align} \frac{(\sqrt[n]{(7^n+n)}-\frac{1}{7})^n}{7^n-n^7}&=\frac{\left(7\left(1+\frac{n}{7^n}\right)^{1/n}-\frac{1}{7}\right)^n}{7^n\left(1-\frac{n^7}{7^n}\right)}\\\\ &=\frac{7^n\left(1+\mathcal{O}\left(\frac{1}{7^n}\right)-\frac{1}{49}\right)^n}{7^n\left(1-\frac{n^7}{7^n}\right)}\\\\ &=\frac{\left(1+\mathcal{O}\left(\frac{1}{7^n}\right)-\frac{1}{49}\right)^n}{\left(1-\frac{n^7}{7^n}\right)}\\\\ &=\frac{\left(\frac{48}{49}\right)^n\left(1+\mathcal{O}\left(\frac{1}{7^n}\right)\right)^n}{1-\frac{n^7}{7^n}}\\\\ &=\frac{\left(\frac{48}{49}\right)^n\left(1+\mathcal{O}\left(\frac{n}{7^n}\right)\right)}{1-\frac{n^7}{7^n}}\\\\ & \sim \left(\frac{48}{49}\right)^n \end{align} $$ and the desired limit is equal to $0$.

$\endgroup$
1
$\begingroup$

When $n \rightarrow \infty$, $7^n >> \mathcal{O}(n^7)>> \mathcal{O}(n)$. Thus,

$$\lim_{n \to \infty} \frac{(\sqrt[n]{(7^n+n)}-\frac{1}{7})^n}{7^n-n^7} = \lim_{n \to \infty} \frac{(7-\frac{1}{7})^n}{7^n} = 0.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.