solution without using the L'Hopital

Question

I want to complete the solution without using the L'Hopital. I got here

$\lim _{x \rightarrow \infty} \frac{\log \left(x^{4}+1\right)+1}{\log \left(x^{3}+2\right)-3}$

$\lim _{x \rightarrow \infty}\left(\frac{\log \left(x^{4}\left(1+\frac{1}{x^{4}}\right)\right)+1}{\log \left(x^{3}\left(1+\frac{2}{x^{3}}\right)\right)-3}\right)$

$\lim _{x \rightarrow \infty}\left(\frac{\log \left(x^{4}\right)+\log \left(1+\frac{1}{x^{4}}\right)+1}{\log \left(x^{3}\right)+\log \left(1+\frac{1}{x^{3}}\right)-3}\right)$

Answer 1

Your are almost done :

$\lim _{x \rightarrow \infty}\left(\frac{\log \left(x^{4}\right)+\log \left(1+\frac{1}{x^{4}}\right)+1}{\log \left(x^{3}\right)+\log \left(1+\frac{1}{x^{3}}\right)-3}\right)$

$= \lim _{x \rightarrow \infty} \frac{4log(x)}{3log(x)}$

$=\frac{4}{3}$