Limit of the sequence $\lim_{n \to \infty}{\frac{\log(n^3-7)}{\sqrt{n}+2/\sqrt{n}}}$ Is there a way mathematically to prove this limit without using de l'hopital because I think I can't use it on sequences. I know I can say that the log function goes to infinity slower than the denominator but is that the only way I can prove this?
$$\lim_{n \to \infty}{\frac{\log(n^3-7)}{\sqrt{n}+2/\sqrt{n}}=0}$$

Let $a_n = \frac{\log(n^3-7)}{\sqrt{n}+2/\sqrt{n}}$. Then
$$0 \leq a_n \leq \frac{\log(n^3)}{\sqrt{n}} = 3\frac{\log(n)}{\sqrt{n}}.$$
So we may instead show that $\lim_{n\to \infty} b_n=0$ where
$$b_n = \frac{\log(n)}{\sqrt{n}}.$$ To show that $\lim_{n\to \infty} b_n=0$ note that for any $0 < p < 1$ we can bound
$$\log(n) = \int_1^n \frac{1}{x} \, dx \leq \int_1^n \frac{1}{x^p}\, dx = \frac{1}{1-p} (n^{1-p} - 1) \leq \frac{1}{1-p}n^{1-p}.$$
Setting, for example, $p=\frac34$, we have
$$\log(n) \leq 4 \sqrt[4]{n},$$
and so
$$0\leq b_n = \frac{\log(n)}{\sqrt{n}} \leq \frac{4}{\sqrt[4]{n}},$$
from which it follows that $\lim_{n\to \infty} b_n=0$.
