Question

lim x-> -infinity x^2e^x

Answer 1

2*e^(-infinity) = 2*e^(1/infinity) = 2e^0 = 2
-infinity^2 = infinity

Answer 2

get it @juliweaver ?

Answer 3

not really sure

Answer 4

do you understand 1/-infinity = 0?

Answer 5

do you understand e^(-x) = e^(1/x)

Answer 6

how are you getting to 1/-infinity? Is it because e raised to any power is 0?

Answer 7

I get your last statement yes

Answer 8

ok if x^(2e^(-x)) = x^(2e^(1/x)) so if we plug in -infinity we get
(-infinity)^(2e^(1/-infinity)
1/-infinity = 0 so we have
(-infinity)^(2*e^0) = (-infinity)^(2*1) = (-infinity)^2

Answer 9

okay their should be a space x^2 e^x

Answer 10

to the negative infinity

Answer 11

\[x^{2} * e^{x} ?\]

Answer 12

yes

Answer 13

so using l'hopitals rule how do you solve?

Answer 14

why?

Answer 15

one sec

Answer 16

okay

Answer 17

would it be -infinity?