Limit and Integral problem work verification-2 I have to calculate the following:

$$\large\lim_{x \to \infty}\left(\frac {\displaystyle\int\limits_{x^{2}}^{2x}t^{4}e^{t^{2}}dt}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)$$

My attempt:
Let $F(x)=\displaystyle\int\limits_0^xt^4e^{t^2}dt$. Then,
$$\large\lim_{x\to\infty}\left(\frac{\displaystyle\int\limits_{x^{2}}^{2x}t^{4}e^{t^{2}}dt}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)=\lim_{x \to \infty}\left(\frac {F(2x) - F(x^2)}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)$$ 
Applying L'Hôpital's rule, we have,
$$\large\begin{align}\lim_{x \to \infty}\left(\frac {32x^4e^{4x^2} - 2x^9e^{x^4}}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}}\right) &= \lim_{x \to \infty}(32x^4e^{4x^2-x} - 2x^9e^{x^4-x}) \\&= \lim_{x \to \infty}\bigg(2x^4e^{4x^2-x}(16-x^5e^{x^4-4x^2})\bigg) = -\infty\end{align}$$
Am I right?
 A: Hint: I strongly suggest the use of power series. We can write down the power series for $e^w$, substitute $t^2$, and integrate term by term to get a series for the top. The series for the bottom is easy to write down.
Remark: As mentioned in a comment, there is an error in the L'Hospital's Rule calculation. It is fixable.
A: Hint: Notice that $$\left(\int_{x^2}^{2x}t^4 e^{t^2} dt\right)' = \color{red}2\dot\,(2x)^4 e^{(2x)^2} - \color{red}{2x}\dot\,(x^2)^4 e^{(x^2)^2}$$
and $$e^x = \sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$$
A: For large $x, x^2 > 2x.$ So it seems like a good idea to write the numerator as $-\int_{2x}^{x^2}t^4e^{t^2}\,dt.$ Notice that the integrand is positive and increasing. Thus the absolute value of the numerator is $\ge (2x)^4 e^{4x^2}(x^2-2x).$ The denominator is $< e^x$ for large $x.$ So in absolute value, it is clear, nay blatantly obvious, that our expression in absolute value $\to \infty.$ Now put the minus sign in to see the limit in question is $-\infty.$
