Find the limit of $a_n=\left(\frac{3n^4+n^2+n(-1)^n}{2n^4+i^n+n^2}\right)_n$
I assume that it tends to $1.5$ since I did some tests. For example $a_{10000}\approx 1.5$.
I also know that $n^4$ increases the fastest and the quotient of their coefficients is equal to $1.5$. But I haven't proved that yet, is there an easier method?
Use the fact that\begin{align}\lim_{n\to\infty}\frac{3n^4+n^2+n(-1)^n}{2n^4+i^n+n^2}&=\lim_{n\to\infty}\frac{\displaystyle3+\frac1{n^2}+\frac{(-1)^n}{n^3}}{\displaystyle2+\frac{i^n}{n^4}+\frac1{n^2}}\\&=\frac{\displaystyle\lim_{n\to\infty}3+\frac1{n^2}+\frac{(-1)^n}{n^3}}{\displaystyle\lim_{n\to\infty}2+\frac{i^n}{n^4}+\frac1{n^2}}.\end{align}