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?
$$\lim_{n \to \infty}{\frac{\log(n^3-7)}{\sqrt{n}+2/\sqrt{n}}=0}$$

Another way to show that ${\log n\over \sqrt{n}}$ tends to $0,$ in order to finish the answer by @Mark Viola,  could be as follows. Let $16^k\le n< 16^{k+1}.$ Then $$\log n<(k+1)\log 16,\quad \sqrt{n}\ge 4^k$$ Hence $${\log n\over \sqrt{n}}\le \log 16\,{k+1\over 4^k}\le {4\log 2 \over 2^k} {k+1\over 2^k}\le {8\log 2 \over 2^{k+1}}\le {8\log 2 \over \sqrt[4]{n}}$$
