Problem

Find the limit. Use l'Hospital's Rule where appro…

02:11

Problem 47 Hard Difficulty

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} $

Answer

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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Video Transcript

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.

Amrita B.

Topics

Derivatives

Differentiation

Volume

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

Problem