Question

is (infinity/0) an indeterminate quotient? where we can apply L'Hopital's rule?

Answer 1

yes, quite indeterminate.

Answer 2

Yeah it's indeterminate quotienent
We can apply in form of limits such as
\[\infty / \infty and 0/0\]

Answer 3

L'Hopital is about limits of differentialble functions.

Answer 4

"Differentialble." Great. I ken spel.

Answer 5

can you use L'H rule for a infinity/0 problem?

Answer 6

I thought it had to be 0/0 or infinity/infinity for L'H rule to apply

Answer 7

No, I don't think so, it wouldn't help except to confirm an undefined limit.

Answer 8

you cant use L'hopitals rule on those...and anywhay ... why would you want to

Answer 9

heres my thing...lim as x-->0 of (cotx-1)/x

Answer 10

if it is \[\infty/0\] then there are only 3 possibilities


either the limit is infinity,-infinity, or does not exits

Answer 11

The answer is suppose to be 0, but I got infinity lol. let me try again

Answer 12

thanks guys!

Answer 13

it is not 0

Answer 14

it is infinity

Answer 15

it was originally lim as x-->0 (cotx -1/x) it was a limit of differences so i had to change it up

Answer 16

i think i changed it wrongly?

Answer 17

cot(x)-(1/x) as x->0 is zero

Answer 18

(cot(x)-1)/x is infinity as x-> 0

Answer 19

wait so i don't have to change it up to apply "L'HR"?

Answer 20

i changed it to (sinx-1)/(xcosx) which look more like something we could use L'H Rule

Answer 21

\[\frac{x\cos(x)-\sin(x)}{x\sin(x)}\]

Answer 22

oh dammit! i mixed up tan with cot...thanks lol!! but wait so if i get something like 0/infinity or infinity/0 that means i changed up the function wrong right? and have to have a better means to change it to apply L'H Rule?

Answer 23

0/infinity or infinity/0 are possible...but dont require l'hospitals rule


\(\frac{0}{0}\) or \(\frac{\pm\infty}{\pm\infty}\)

to use the rule

Answer 24

http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

Answer 25

alright thanks everyone!!!