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# Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.$\displaystyle \lim_{x\to \infty} x^3 e^{-x^2}$

## This limit has the form $\infty \cdot 0 . \lim _{x \rightarrow \infty} x^{3} e^{-x^{2}}=\lim _{x \rightarrow \infty} \frac{x^{3}}{e^{x^{2}}} \stackrel{\underline{H}}{=} \lim _{x \rightarrow \infty} \frac{3 x^{2}}{2 x e^{x^{2}}}=\lim _{x \rightarrow \infty} \frac{3 x}{2 e^{x^{2}}} \underline{\underline{H}} \lim _{x \rightarrow \infty} \frac{3}{4 x e^{x^{2}}}=0$

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first things first. Let's rewrite this without the negative exponents. It's easier to read now. We know that we have number and denominator both go towards infinity as X goes towards infinity. Therefore, we can apply the hospital's rule. Therefore, take the derivative of top and bottom three X squared over two acts each. The X word you know you can cross off one X Therefore, we have three acts on the top over to eat of the X squared. Now we capture a pile hospital's rule again In order to actually get an answer, we end up with three over infinity. Remember anything divide by infinity is zero.

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