Question

L'Hopitals' Rule - can anyone check my answer?

lim x-> inf of (x^2) / (ln x)

inf / inf form, so
lim x->inf of 2x^2 and
lim x-> inf of 2(inf)^2 = inf <--final answer.

Is this right?

Answer 1

it is in that form, but you should do it in your head

Answer 2

log grows very very very slowly, much slower than any polynomial

Answer 3

I should do it in my head?

Answer 4

i don't understand.. so where did i get it wrong?

Answer 5

but yes, if you use l'hopital you get the right answer
i just mean you should know in advance what the answer is

Answer 6

you didn't do it wrong, you did it right

Answer 7

oh... well but i don't know... haha

Answer 8

log grows slower than \(x\) or \(x^2\) so
\[\lim_{x\to \infty}\frac{x^2}{\log(x)}=\infty\] for sure

Answer 9

think if it was base ten (which it isn't but might as well be)
if you had \(x=1,000,000\) then it would be
\[\frac{1,000,000,000,000}{6}\] a pretty big number!

Answer 10

i dont think i get infinite limits in the first place... wow...

Answer 11

so i''m ust following the step by step directions for finding these limits with l'hopital rule...

Answer 12

Thank you again @satellite73

Answer 13

you will develop intuition as you go along

Answer 14

thank you @thomasz ! but i did the same work you did to find the answer :)

Answer 15

You too! :D @thomasz