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Use the L'Hopital rule to solve:

$$\lim_{x\to-\infty}x^2\cdot e^x$$

I need a quotient of infinities or of zeroes. One way could be this:

$$\frac{e^x}{x^{-2}} = \frac{0}{0}$$

So we apply the L'Hopital rule:

$$\frac{e^x}{-2x^{-3}}$$

Oh, this is not going to work. We'll be applying L'Hopital countless times without luck.

Perhaps this arrangement will do:

$$\frac{x^{2}}{e^{-x}} = \frac{\infty}{0}$$

Nope.

Maybe I could use one of the properties of natural logarithms. Since $e^x = \ln(x)$ I could have

$$\frac{x^2}{\ln(-x)} = \frac{\infty}{\infty}$$

Great! Now we can apply the L'Hopital rule:

$$\frac{2x}{\frac{-1}{\ln(-x)}} = \frac{-\infty}{0}$$

Dammit. Well, maybe we can re-arrange this:

$$\frac{2x}{\frac{-1}{\ln(-x)}} = 2x\cdot -\ln(-x)$$

Then,

$$2x\cdot -\ln(-x) = -\infty \cdot -\infty = \infty$$

Apparently this is wrong. The answer should be $0$.

What was my mistake?

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  • $\begingroup$ Hint : try the substitution $y = -x$ . $\endgroup$ – Victor Nov 9 '15 at 21:10
  • $\begingroup$ how do you figure that $e^x = ln(x)$ ? $\endgroup$ – costrom Nov 9 '15 at 21:12
  • $\begingroup$ @Zol I think you can make a generalization of $x^ne^x$ when $x$ goes to negative infinity? $\endgroup$ – imranfat Nov 9 '15 at 22:22
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The nope should be a yeah! $$ \lim_{x\to-\infty}e^{-x} = \lim_{x\to \infty}e^{x} = \infty $$

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  • 1
    $\begingroup$ Just so this is clear to the original poster, notice that $x$ is going to negative infinity. $\endgroup$ – user43395 Nov 9 '15 at 21:19
  • $\begingroup$ Do we have to change $x\to -\infty$ to $x \to \infty$? Can't we just keep $e^{-x}$? Because when you evaluate, it will become $e^{--\infty}$ anyway. $\endgroup$ – Zol Tun Kul Nov 9 '15 at 21:22
  • $\begingroup$ No you do not have to change directions. I just thought that made the limit value a bit easier to understand. $\endgroup$ – mvw Nov 9 '15 at 21:24
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$$\lim_{x\to-\infty}x^2\cdot e^x=\lim_{x\to\infty}\frac{x^2}{ e^x}$$

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$$\lim_{x\to -\infty}e^xx^2=$$ $$\lim_{x\to -\infty}\frac{x^2}{e^{-x}}=$$ $$\lim_{x\to -\infty}\frac{\frac{\text{d}}{\text{d}x}x^2}{\frac{\text{d}}{\text{d}x}e^{-x}}=$$ $$\lim_{x\to -\infty}\frac{2x}{-e^{-x}}=$$ $$\lim_{x\to -\infty}-2e^xx=$$ $$-2\left(\lim_{x\to -\infty}e^xx\right)=$$ $$-2\left(\lim_{x\to -\infty}\frac{x}{e^{-x}}\right)=$$ $$-2\left(\lim_{x\to -\infty}\frac{\frac{\text{d}}{\text{d}x}x}{\frac{\text{d}}{\text{d}x}e^{-x}}\right)=$$ $$-2\left(\lim_{x\to -\infty}\frac{1}{-e^{-x}}\right)=$$ $$-2\left(\lim_{x\to -\infty}-e^x\right)=$$ $$2\left(\lim_{x\to -\infty}e^x\right)=$$ $$2\left(\exp\left(\lim_{x\to -\infty}x\right)\right)=0$$

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$ln(e^x) = x$, not $ln(x) =e^x$ And indeed, the limit is $0$.

You have $x^2 e^x = \frac{x^2}{e^{-x}}$, and, by applying l'Hôpital's rule twice, you get :

$\lim_{x\rightarrow-\infty} x^2 e^x = \lim_{x\rightarrow-\infty} \frac{2x} {-e^{-x}} = \lim_{x\rightarrow-\infty} \frac{2} {e^{-x}} = \lim_{x\rightarrow-\infty} 2 e^x = 0 $

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