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I have used L'Hospital's Rule on $\lim_{x \to \infty} x^3e^{-x^2}$ to get $\lim_{x \to \infty} 3x^2e^{-x^2}-2x^4e^{-x^2}$. I plugged the latter form of the limit into desmos.com and found that as $x$ approaches infinity, the limit equals zero.

$$\lim_{x \to \infty} x^3e^{-x^2} = \lim_{x \to \infty} \frac{d}{dx} x^3e^{-x^2} = \lim_{x \to \infty} [3x^2 \cdot e^{-x^2}]+[x^3 \cdot -2xe^{-x^2}] = \lim_{x \to \infty}3x^2e^{-x^2}-2x^4e^{-x^2}$$

However, I can't rely on desmos in an exam, and plugging in multiple values for x takes a long time. Is there an easier way to solve limits to infinity other than plugging in multiple values for $x$?

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You use L'Hospital's rule only when you have indeterminate expressions of the form $\frac{\pm\infty}{\pm\infty}$ or $\frac{0}{0}$ (all other possible indeterminate forms, such as $0\cdot \infty$, $0^\infty$ etc, must be converted to those two before applying the rule):

$$ \lim_{x \to \infty} x^3e^{-x^2} =\lim_{x \to \infty} \frac{x^3}{e^{x^2}}\ \left(\infty\cdot 0\rightarrow\frac{\infty}{\infty}\right) $$

This is an example of the $\frac{\infty}{\infty}$ indeterminate form. So, use L'Hospital's rule by taking the derivative of the top and then of the bottom and then possibly do it again (always make sure what you have is still one of those two indeterminate forms) until you reach an expression that you can finally take the limit of.

$$ \lim_{x \to \infty} x^3e^{-x^2} =\lim_{x \to \infty} \frac{x^3}{e^{x^2}} \stackrel{\text{L'H}}{=} \lim_{x \to \infty} \frac{3x^2}{2xe^{x^2}} = \lim_{x \to \infty} \frac{3x}{2e^{x^2}}\stackrel{\text{L'H}}{=} \lim_{x \to \infty} \frac{3}{4xe^{x^2}}=0 $$

And it only makes sense that this limit is zero. The exponential $e^{x^2}$ on the bottom as $x$ goes to infinity is sure going to overwhelm the polynomial $x^3$ on the top making the fraction go to zero.

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Hint: The limit is equal to

$$\lim\limits_{x\to\infty} \frac{x^3}{e^{x^2}} \stackrel{\text{L'Hospital}}{=} \lim\limits_{x\to\infty}\frac{3x^2}{2xe^{x^2}} \stackrel{\text{L'Hospital}}{=} \cdots$$

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You can only use l'Hopital’s rule on functions that evaluate to two indeterminate forms $(\frac 00, \frac{\infty}{\infty})$. The form you had $\frac{x^3e^{-x^2}}{1}$ is not in the correct indeterminate form, so you should make it into the correct form before differentiating.

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  • $\begingroup$ This is incorrect. The given question is in the indeterminate form of $\infty \times 0$. One can only apply L'Hôpital's rule for the specific indeterminate forms - $\frac{\pm\infty}{\pm\infty}$ and $\frac00$. $\endgroup$ – Peter Foreman Apr 5 at 20:24
  • $\begingroup$ @PeterForeman oops sorry! $\endgroup$ – D.R. Apr 5 at 20:29

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