Question

lim x->0+ (lnx)squared/ ln(sinX)

Answer 1

you know l'hospital's rule?

Answer 2

yep thats what im lookin at but havin a hard time figuring out limit at 0+

Answer 3

approaching 0+***

Answer 4

keeps hangin me up on other problems also so wanna understand using this problem ... hopefully

Answer 5

i believe the limit is neg infinity

you have to use L'hopitals rule as mentioned above
then after taking derivatives and simplifying fraction
use known limit
\[\lim_{x \rightarrow 0}\frac{\sin x}{x} = 1\]
that will make it possible to evaluate limit as x->0+

Answer 6

so first indeterminate form would be 0/0?

Answer 7

actually for this case its infinity/infinity

Answer 8

either way its indeterminate

Answer 9

sinx as x->0+ would be inifinity and ln of of this is inifinity also?

Answer 10

yea i do kno its indeterminate

Answer 11

just having a hard time vizualizing... feel like im havin a major brainfart

Answer 12

lim of ln(0) is neg infinity right?

sin(0) = 0

so both top/bottom are infinity

Answer 13

yes this is true

Answer 14

thanks

Answer 15

yw