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.
 A: 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 $$\lim_{x\to \infty}{\frac{120x}{24}}=\infty$$
A: As a side note :
Unless the sign is obvious like in $\sum\limits_{i=1}^{\infty}$ get used to explicitely write the sign of infinity $+\infty$ and $-\infty$, especially with the limit operator. 
Without signing, you consider implicitely consider the two limits in $\pm\infty$. Here it happens that $\frac1x$ has the same limit $0$ in $+\infty$ and in $-\infty$ so it doesn't matter.
Yet that explains why $\frac10$ is indeterminate :


*

*if $x\to0^+$ that is $x\to0$ and $x\ge0$ then $\frac1x\to+\infty$

*if $x\to0^-$ that is $x\to0$ and $x\le0$ then $\frac1x\to-\infty$


Thus $\frac10$ has two limits and this is the definition of being indeterminate.
But if you write $\displaystyle{x=\frac1{\frac 1x}}$ then $x\to+\infty$ transforms to $\frac1{0^+}$ which is not indeterminate anymore.
Same to the other side.
A: First off, division by $0$ is undefined. Try this, $$\lim_{x\to\infty}\frac{x^5+x^3+4}{x^4-x^3+1}=\lim_{x\to\infty}\frac{x+\frac{1}{x}+\frac{4}{x^4}}{1-\frac{1}{x}+\frac{1}{x^4}}=\infty$$
This shows more clearly the limit without resulting in $\frac{1}{0}$
A: 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.
