L'Hôpital's Rule and Infinite Limits I was wondering if anyone could help me with computing a limit using L'Hôpital's Rule.
Using L'Hôpital Rule for the following limit, I get the following result:
\begin{equation}
\lim_{x \to 0} \frac{e^x-1-x^2}{x^4+x^3+x^2} \therefore \lim_{x \to 0} \frac{e^x-1-x^2}{x^4+x^3+x^2} \stackrel{L'H}{=} \lim_{x \to 0} \frac{e^x-2x}{4x^3+3x^2+2x}
\end{equation}
(1/0) which means that the limit is either $+\infty$, $-\infty$ or it does not exist.
If I want to find out the answer, I should take the side limits - but here lies my question: the side mimits of what? The original function or the function after L'Hôpital rule? Or any of the two?
Should I do this
\begin{equation}
\lim_{x \to 0^-} \frac{e^x-2x}{4x^3+3x^2+2x} = -\infty
\end{equation}
\begin{equation}
\lim_{x \to 0^+} \frac{e^x-2x}{4x^3+3x^2+2x} = +\infty
\end{equation}
Or this?
\begin{equation}
\lim_{x \to 0^-} \frac{e^x-1-x^2}{x^4+x^3+x^2} = -\infty
\end{equation}
\begin{equation}
\lim_{x \to 0^+} \frac{e^x-1-x^2}{x^4+x^3+x^2} = +\infty
\end{equation}
Coincidentally (or not) both limits (from the right of both equations and from the left of both equations) give the same answer. Therefore, the limit does not exist.
For the other function below, the same thing happens:
\begin{equation}
\lim_{x \to 0} \frac{sin(x)}{x^4+x^3}
\end{equation}
So what I dind't find out is which function I should use, or if it really doens't matter.
Thank you

The right way to proceed is the following from what you have found
$$\lim_{x \to 0^+} \frac{e^x-2x}{4x^3+3x^2+2x} = \infty \implies \lim_{x \to 0^+} \frac{e^x-1-x^2}{x^4+x^3+x^2} = \infty$$
$$\lim_{x \to 0^-} \frac{e^x-2x}{4x^3+3x^2+2x} = -\infty \implies \lim_{x \to 0^-} \frac{e^x-1-x^2}{x^4+x^3+x^2} = -\infty$$
where the LHS are evaluated by l'Hospital's rule and the RHS are derived form this result.
Therefore in this case we conclude that limit doesn't exist.
For the other one we can also proceed as follows
$$\frac{\sin(x)}{x^4+x^3}=\frac{\sin(x)}{x}\frac{x}{x^4+x^3}=\frac{\sin(x)}{x}\frac{1}{x^3+x^2} \to 1 \cdot \infty=\infty$$
