L'Hospital's Rule: Find Limit of x^2e^x

[endeavor]

Homework Statement

Find the limit.
\lim_{x\rightarrow -\infty} x^2e^x

Homework Equations

L'Hospital's Rule.

The Attempt at a Solution

I rewrite the limit:
\lim_{x\rightarrow -\infty} \frac{e^x}{\frac{1}{x^2}}
applying L'Hospital's Rule:
\lim_{x\rightarrow -\infty} \frac{e^x}{-2(\frac{1}{x^3})}
But the numerator and denominator still go to zero...If I keep applying L'Hospital's Rule, I still won't get anywhere. It looks like x's exponent will continue to grow; then if I put the x term back in the numerator, it would seem like the x term would overcome e^x, and the limit would go to infinity. However, I graphed the function, and it seems to go to zero. What am I doing wrong?

 

[Curious3141]

Hint : try rewriting the limit as \lim_{x\rightarrow \infty} {(-x)}^2e^{-x} = \lim_{x\rightarrow \infty} \frac{x^2}{e^x} and apply L'Hopital's rule twice.

 

[Hurkyl]

it would seem like the x term would overcome e^x, and the limit would go to infinity.

Let this be your first lesson that exponential factors dominate polynomial factors.

Anyways, as Curious suggests, try moving the exponential to the bottom instead of the monomial.

 

[endeavor]

Yeah, that works. It must've slipped my mind that \lim_{x\rightarrow -\infty} f(x) = \lim_{x\rightarrow \infty} f(-x).

Is there any other way to do it though?

 

[theperthvan]

intuition.
Looking at this one
\lim_{x\rightarrow \infty} \frac{x^2}{e^x}

Which one approaches infinity 'faster'?

 

[HallsofIvy]

For any positive n, a> 1, a^x goes to infinity faster than xⁿ so the fraction you have goes to 0. We say that a^x "dominates" xⁿ as Hurkyl said.


For example, for large enough x, 1.0000001^x is larger than x^100000000.