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

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

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 $0$.
When $n \rightarrow \infty$, $7^n >> \mathcal{O}(n^7)>> \mathcal{O}(n)$. Thus,
$$\lim_{n \to \infty} \frac{(\sqrt[n]{(7^n+n)}-\frac{1}{7})^n}{7^n-n^7} = \lim_{n \to \infty} \frac{(7-\frac{1}{7})^n}{7^n} = 0.$$