Limit of $\frac{1}{7}e^{-2x^2}(1-4x^2)$ as $x\to\infty$ I calculated the derivative of $\frac{x}{7}*e^{-2x^2}$ and got $\frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)
I don't know how to find the limit of this function:
$$\frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$\lim_{x \to \infty} \frac{1}{7}e^{-2x^2}+\lim_{x \to \infty} \frac{1}{7}xe^{-2x^2}*(-4x) $$

Using L'Hôpital's rule , we get 
\begin{align}
\lim_{x\to \infty}\frac{1-4x^2}{7e^{2x^2}}&=\lim_{x\to \infty}\frac{-8x}{28\cdot xe^{2x^2}}\\
&=\lim_{x\to \infty}\frac{-8}{28e^{2x^2}}=0
\end{align}. 
