Question

lim goes to -infiniti, [e^(3x) - 2x] / [(1/x) - x]

Answer 1

2

Answer 2

use l'hospital's rule

Answer 3

[3e^(3x) - 2] / [-1/(x^2) - 1]

analysing the limit you get

- 2 / -1

which is 2

Answer 4

ok, but if you take 3e^3x to -infiniti it equals infiniti, not zero

Answer 5

3x^(-infinity)=3/x^(infinity) = 0

Answer 6

i mean e^(infinity) not x^(infinity)

Answer 7

wow i was totally missing that the e^infinity drops down to the denominator, whoops! thanks guys