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
 A: If you have the possibility you can always make a plot (for example with wolfram alpha) to check if your calculations are correct. I come to the same result for the first limit however for the second one I get $\lim_{x\to 0}\frac{\sin(x)}{x^4+x^3}=\lim_{x\to 0}\frac{\cos(x)}{4x^3+x^2}$, which tends to $+\inf$ for both limits (notice that $x^2$ dominates $x^3$ in this case)
A: 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
$$\frac{\sin(x)}{x^4+x^3}=\frac{\sin(x)}{x}\frac{x}{x^4+x^3}=\frac{\sin(x)}{x}\frac{1}{x^3+x^2} \to 1 \cdot \infty=\infty$$
A: L'Hôpital's Rule is applicable when say f and g are are differentiable functions on open intervals K except at maybe a point 'c'.
Mathematically put,
IF $\lim_{x\to c} f(x) = \lim_{x\to c} g(x) = 0 $ or $\pm\infty$, and 
$ g'(x) \neq 0$ for all x in K except $x\neq c$
THEN $\lim_{x \to c} \frac{f(x)}{g(x)}=\lim_{x \to c} \frac{f'(x)}{g'(x)}$.
In your equation, thus L'Hôpital's Rule can be applied repeatedly to get the answer 1/24.
