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 $$\lim_{x\to \infty}{\frac{120x}{24}}=\infty$$
First of all, $\frac 1 0 = \infty$, meaning if $f(x)\to 1$ and $g(x)\to 0$ (and $g(x)$ is positive) then $\frac {f(x)} {g(x)}\to +\infty$. It's not an indeterminate form.

However, even if it were an indeterminate form, that's not a problem. Obtaining an indeterminate form doesn't mean the limit doesn't exist, it just means "the method you used doesn't work, try something else". For example, the constant function $1$ definitely has a limit as $x\to\infty$, but if I do something crazy like rewrite it as $\frac x x$, then I get the indeterminate form $\frac \infty \infty$. Every limit can be made to look like an indeterminate form if you try to calculate it in the right way.