Limit of this sequence

Question

Hi I am learning for a calculus exam and I can't seem to find a way to find this sequence's limit

$$ a_n = \frac{3\ln(n^2+2n)}{\ln(\sqrt{n^4-1})+3\ln(5\ln(2n))} $$

I tried L'Hopitals rule, but then it gets really messy.

I would really appriciate it, if someone could give me hint.

Thanks.

Answer 1

Rewrite it as $$ a_n=3\frac{2\ln n+\ln(1+\frac{2}{n})}{2\ln n+\ln\sqrt[4]{1-\frac{1}{n^4}}+3\ln(5\ln(2n))}= 3\frac{1+\dfrac{\ln(1+\frac{2}{n})}{2\ln n}}{1+\dfrac{\ln\sqrt[4]{1-\frac{1}{n^4}}}{2\ln n}+\dfrac{3\ln(5\ln(2n))}{2\ln n}} $$