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$$\lim_{x \to \infty} \frac{1}{7}e^{-2x^2}+\lim_{x \to \infty} \frac{1}{7}xe^{-2x^2}*(-4x) $$

Hint: Rewrite the expression as
$$\frac{1}{7}e^{-2x^2}\left(1-4x^2\right) = \frac{1-4x^2}{7e^{2x^2}}$$
Now, notice the growth of the numerator and denominator. Which grows more quickly?
