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

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    $\begingroup$ Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-\infty$. Their product goes to $0.$ $\endgroup$
    – Doug M
    Jan 3, 2019 at 15:45
  • $\begingroup$ Is $*$ multiplication or convolution in your question? $\endgroup$
    – Asaf Karagila ♦
    Jan 4, 2019 at 7:50

5 Answers 5

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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?

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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}.

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    $\begingroup$ L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler. $\endgroup$
    – Andreas Rejbrand
    Jan 4, 2019 at 7:01
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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 $0$

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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 $$ \begin{align} \left|\frac17e^{-2x^2}\!\!\left(1-4x^2\right)\right| &\le\frac{4x^2+1}7\frac1{2x^4}\\ &=\frac2{7x^2}+\frac1{14x^4}\tag3 \end{align} $$

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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.

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  • $\begingroup$ Thank you I completely forgot the L'Hopitals Rule $\endgroup$
    – mashhc
    Jan 3, 2019 at 16:04
  • $\begingroup$ @B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one. $\endgroup$
    – Ethan Bolker
    Jan 4, 2019 at 1:34

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