Question

Use L'Hospital Rule
limx->infinity x^3*e^-x^2

Answer 1

\[\lim_{x \rightarrow \infty} x^{3}*e^{-x^2}\]

Answer 2

\[\lim_{x \rightarrow \infty} x^3*e^{-x^2}\]

Answer 3

yes

Answer 4

i just do not know how to type it in that way when writing the original question

Answer 5

You could differentiate a couple times, since it's indeterminate, and eventually x^3 becomes 0.

Answer 6

\[\lim_{x \rightarrow \infty}\frac{ x^3 }{ e^{x^2} }\]

Answer 7

^ like that.

Just from glancing at it, it'll approach zero (e^x approaches 0 faster than x^3 approaches infinity) - this is a good way to gauge your answer.

Answer 8

how do you know it becomes 0. I thought it would go to infinity becasue I can not get rid of the e^x

Answer 9

exponential functions grow at a faster rate than polynomials. Or in this case, shrink at a faster rate.

Answer 10

yes, i got that first step asaad123

Answer 11

Now use l hopital and differentiate the numerator and denominator.

Answer 12

ooooo, soo smart. geeze, ty

Answer 13

So you can often know the answer before proving it, which helps to check your answer.

affter differenitating once....\[\Large \lim_{x \rightarrow \infty}\frac{ 3x^2 }{ 2x e^{x^2} }\]

Answer 14

so basiaclly because i would set it up for the exp to go in the bottom itll always go to zero regardless of what the num is right?

Answer 15

because nothing grows faster than an exp

Answer 16

Provided your a in a^x is bigger than 1

If it's something like a polynomial over an exponential, yes

\[\Large \frac{ x^9 }{ 1.1^x}\] eventally this should approach zero.

Answer 17

And if you graph it, it does... eventually. https://www.google.com/search?q=x%5E9%2F1.1%5Ex&oq=x%5E9%2F1.1%5Ex&aqs=chrome..69i57.5734j0&sourceid=chrome&espvd=210&es_sm=93&ie=UTF-8 zoom out a lot and you'll see it.

Answer 18

so if i was to keep d/dx it ill eventually get 3/infin right?

Answer 19

which =0

Answer 20

right

Answer 21

TYsoVM :)

Answer 22

\[\Large \lim_{x \rightarrow \infty}\frac{ 3x^2 }{ 2x e^{x^2} }\] simplify it first, \[\Large \lim_{x \rightarrow \infty}\frac{ 3x }{ 2 e^{x^2} }\] now differntiate again.

Answer 23

gotcha