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$$\lim_{x \to \infty} \frac{1}{7}e^{-2x^2}+\lim_{x \to \infty} \frac{1}{7}xe^{-2x^2}*(-4x) $$
 A: 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?
A: 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}. 
A: 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 $0$
A: Note that for $x\ge0$,
$$
\begin{align}
e^x
&=1+x+\frac{x^2}2+\dots\\
&\ge\frac{x^2}2\tag1
\end{align}
$$
Thus,
$$
e^{-2x^2}\le\frac1{2x^4}\tag2
$$
Applying $(2)$ to the expression for the derivative gives
$$
\begin{align}
\left|\frac17e^{-2x^2}\!\!\left(1-4x^2\right)\right|
&\le\frac{4x^2+1}7\frac1{2x^4}\\
&=\frac2{7x^2}+\frac1{14x^4}\tag3
\end{align}
$$
A: You pretty much have the solution.  The following limit: $$\frac{1}{7}\lim_{x\to \infty}\frac{1}{e^{2x^2}}=0.$$
I do not believe that was your problem.  The last bit is what you were probably having trouble with.  I would try u substitution.  Let $u=x^2$ then we have the following:
$$-\frac{4}{7} \lim_{u\to\infty}\frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.
