Limit as $x \to \infty$ of $\frac{x^5+x^3+4}{x^4-x^3+1}$ Suppose we have to find the following limit
$\lim_{x\to\infty}\frac{x^5+x^3+4}{x^4-x^3+1}$
Now, if we work with the De L'Hopital rule with successive differentiations we get $L=+\infty$
But if we work like this instead: $$L=\lim_{x\to\infty}\frac{x^5(1+\frac{1}{x^2}+\frac{4}{x^5})}{x^5(\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^5})}$$ then $L$ does not exist.
What is correct and what is false here? I'm a little confused.

First you should avoid division by 0 when calculating limit, as it will tell nothing about the limit. $\frac{1}{0}$ is not infinity, it is undefined. $$\lim_{x\to 0}{\frac{1}{x}}$$ is different scenario. Here you get limit is infinity. In your example you have to apply L'Hopital's rule $4$ times to get $$\lim_{x\to \infty}{\frac{120x}{24}}=\infty$$
