Indeterminate form and L'Hospital rule 
$$
\lim_{x \to \infty}\frac{e^x}{x^2}
= \lim_{x \to \infty}\frac{e^x}{2x}    
= \lim_{x \to \infty}\frac{e^x}{2}          
= \infty
$$

Can anybody tell me please why L'hospital's rule been used only two times?
From my point of view, $\frac{\infty}{\infty}$ is indeterminate form but at the end, only $\infty$ is also indeterminate form.
 A: Your application is correct indeed also $\frac{e^x}{2x}$ is in the form $\frac{\infty}{\infty}$ while the last one is not an indeterminate form and the limit is indeed infinity.
As an alternative note that eventually $e^x\ge x^3$ and then
$$\frac{e^x}{x^2}\ge \frac{x^3}{x^2}\ge x\to \infty$$
We can prove that also by induction, that is $e^n \ge n^3$


*

*base case: $n=1 \implies e\ge 1$, $n=2 \implies e^2\not\ge 8$, $n=3 \implies e^3\not\ge 27$, $n=4 \implies e^4\not\ge 64$,$n=5 \implies e^5\ge 125$

*induction step: assume true $e^k \ge k^3$ for some $k$ then


$$e^{k+1}=e\cdot e^k \ge e\cdot k^3\ge 2\cdot k^3\stackrel{?}\ge (k+1)^3$$
and the latter is true indeed
$$2\cdot k^3\ge (k+1)^3$$
$$k(k^2-3k-3)\ge 1, \quad k\ge 4$$
A: Note that $\infty$ is not an intermediate form. Simply put :
$$\lim_{x \to \infty}  \frac{e^x}{2} = \frac{1}{2}\lim_{x \to \infty} e^x =\infty$$
Indeterminate forms are typically considered to be :
$$\frac{0}{0}, \frac{\infty}{\infty}, 0 \cdot \infty, 1^\infty, \infty  - \infty, 0^\infty,\infty^0$$
