Question

L-Hopital's Rule: lim x--> infinity of (e^3x)/(x^2)

Answer 1

what happens when x approaches infinity? what form does (e^3x)/(x^2) take on?

Answer 2

it is indeterminate: infinity/infinity

Answer 3

correct, so `L-Hopital's Rule` does apply here

Answer 4

I'm assuming the numerator is \(\LARGE e^{3x}\) right?

Answer 5

yes

Answer 6

what is the derivative of e^(3x) equal to?

Answer 7

e^3x

Answer 8

you forgot to apply the chain rule

Answer 9

you need to derive the inner stuff (3x) to get 3

so if y = e^(3x) then dy/dx = 3*e^(3x)

Answer 10

oh right!

Answer 11

now take the derivative of the denominator x^2 to get ______

Answer 12

2x

Answer 13

then e^infinity / infinity)

Answer 14

so we go from
\[\Large \lim_{x \to \infty} \left[\frac{e^{3x}}{x^2}\right]\]
to
\[\Large \lim_{x \to \infty} \left[\frac{3e^{3x}}{2x}\right]\]

Answer 15

if that second limit leads to infinity/infinity, you'll need to apply `L' Hospital's Rule` again

Answer 16

so: e^3x (0) / 2

Answer 17

as the derivative again?

Answer 18

you should go from
\[\Large \lim_{x \to \infty} \left[\frac{3e^{3x}}{2x}\right]\]
to
\[\Large \lim_{x \to \infty} \left[\frac{9e^{3x}}{2}\right]\]

Answer 19

I don't think you applied the chain rule properly

Answer 20

oh you're right! i see my mistake

Answer 21

you get infinity / 2

Answer 22

so the answer is infinity?

Answer 23

anyways at this point, you should find that the entire expression now approaches infinity as x approaches infinity

ie,

\[\Large \lim_{x \to \infty} \left[\frac{9e^{3x}}{2}\right] = \infty\]

Answer 24

yep,

\[\Large \lim_{x \to \infty} \left[\frac{e^{3x}}{x^2}\right] = \infty\]

Answer 25

thank you so much! this really helped!!

Answer 26

no problem