Question

Question #8536a

Answer 1

The limit is `0`.

Explanation:

`lim_(xrarr-oo) x^2e^(8x)` has indeterminate initial form `oo * 0`

Rewrite it to use l'Hospital's Rule

`lim_(xrarr-oo) x^2/e^(-8x)` Has intial form `oo/oo` so we can apply l'Hospital.

`(d/dx(x^2))/(d/dx(e^(-8x))) = (2x)/(-8e^(-8x))` which has form `-oo/-oo` so use llHospital again.

`(d/dx(2x))/(d/dx(-8e^(-8x))) = 2/(64e^(-8x)) = 1/32e^(8x)` which goes to `0` as `xrarr-oo`