Question

limit (x^2-x^3)e^2x
x-->-infinity

Answer 1

this is what i got so far

Answer 2

(x^2-x^3)/e^-2x

Answer 3

i later did the derivative which is (2x-3x^2)/-2e^-2x

Answer 4

is it right?

Answer 5

my book says that the answer is 0

Answer 6

So you were able to get an indeterminate form of \(\Large\rm \frac{\infty}{\infty}\) which allowed you to apply L'Hospital's Rule?

Yes, looks good so far :)
Derivative is correct.

Answer 7

i am having trouble find the answer

Answer 8

What form are you getting now?
Is it still indeterminate?

Answer 9

yes so now its 2-6x/4e^-2x

Answer 10

right?

Answer 11

Good good good.
Notice that the x's are deteriorating while the exponential is super strong. Can't touch that guy :)

So you will EVENTUALLY get to a point where only the exponential is left.
You might have to L'Hop one more time.

Answer 12

so then i got -6/-8e^-2x

Answer 13

Just make sure you're checking the form between each application of L'Hop.
It has to stay in the form \(\Large\rm \frac{0}{0}\) or \(\Large\rm \frac{\infty}{\infty}\)

Answer 14

mmm k

Answer 15

so when i plug in the -infinity times the -2 do i get positive infinity?

Answer 16

Don't let your teacher hear you say "plug in -infinity", that will annoy them :) lol

But yes, it looks like we're approaching this form,\[\Large\rm \frac{6}{8\cdot \infty}\]Right?
So what is the expression overall approaching?

Answer 17

oh sorry so it would be 6/infinty and that equals 0 right?

Answer 18

yes, the limit appears to be approaching zero overall.

Good job \c:/

Answer 19

yes thank you