Limit of $\frac{1}{7}e^{-2x^2}(1-4x^2)$ as $x\to\infty$

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?

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}.

Your derivative is correct.

Then arrange it as:

$$\frac{1-4x^2}{7e^{2x^2}}$$

Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$