When to Stop Using L'Hôpital's Rule

Once your answer is no longer in the form 0/0 or $\frac{\infty}{\infty}$ you must stop applying the rule. You only apply the rule to attempt to get rid of the indeterminate forms. If you apply L'Hopital's rule when it is not applicable (i.e., when your function no longer yields an indeterminate value of 0/0 or $\frac{\infty}{\infty}$) you will most likely get the wrong answer.

You should have stopped differentiating the top and bottom once you got to this:

$\dfrac{e^x-2x}{4x^2+3x^2+2x}$. Taking the limit at that gives you $1/0$. The limit is nonexistent.

Also, don't be tempted to say "infinity" when you see a 0 in the denominator and a non-zero number in the top. It may not be the case. For example, the function $\frac{1}{x}$ approaches infinity and negative infinity from both sides of the limit as x approaches 0. Its not necessarily infinite; its best just to leave it as "nonexistent".

After differentiating just once, you get $$\lim_{x \to 0} \dfrac{e^x-2x}{4x^3+3x^2+2x}$$ which "evaluates" to $\dfrac 10$, i.e., the numerator approaches $1$, and the denominator approaches $0$. Hence, L'Hopital no longer applies and we have $$\lim_{x \to 0} \dfrac{e^x-2x}{4x^3+3x^2+2x}\quad\text{does not exist}.$$

L'Hopital's rule applies provided and only while a limit evaluates to an "indeterminate" form: e.g., $\dfrac 00, \;\text{or}\;\dfrac {\pm\infty}{\pm\infty}$.

A quick addition to Ra1nMaster's otherwise excellent answer: you can only apply L'Hopital's rule if you have an indeterminate form and if the limit, after applying L'Hopital's rule, exists.

This second condition is equally important; for instance a classic stumper is $$\lim_{x\to\infty} \frac{x}{x+\sin x}.$$ Since this limit has the form $\frac{\infty}{\infty}$, one might naively apply L'hopital's rule, getting $$\lim_{x\to\infty} \frac{1}{1+\cos x}$$ and concluding the original limit does not exist. This is wrong; $$\lim_{x\to\infty} \frac{x}{x+\sin x} = \lim_{x\to\infty} \frac{1}{1+\frac{\sin x}{x}}=1.$$