- #1

IntegrateMe

- 217

- 1

^{3}or 2x

^{3}+ ln(x)?

**Steps:**

**1)**lim (x-4)

^{3}/[2x

^{3}+ ln(x)]

**2)**The limit is infinity/infinity as x tends to infinity, so we can use L'H rule.

**3)**Applying L'H rule we get [3x(x-4)

^{2}]/(6x

^{3}+ 1)

**4)**The limit is again infinity/infinity as the above function tends to infinity, so we apply L'H again.

**5)**Applying L'H rule we get [3(x-4)(3x-4)]/18x, which yet again yields infinity/infinity as x tends to infinity, so, we apply L'H rule again.

**6)**Applying L'H rule we get (18x-48)/18 which is equal to infinity/18 = infinity.

Solution:

(x-4)

^{3}grows faster because the limit approaches infinity. I know i did something wrong because the answer is that

**they grow at the same rate**. Can anyone point out my mistake(s)? Thanks!