Question

is this correct? using l'hopitals rule
limit as x goes to 1+
ln(x^2)/(x^2-1)=1

Answer 1

first off start with
\[\frac{2\ln(x)}{x^2-1}\]

Answer 2

this will make your differentiation much easier
then you should be good to go

Answer 3

First confirm that it's indeterminate form. Evaluating at x=1 gives 0/0. Cool, we can use L'hopital's! Take the derivative on top and on bottom gives:
\[\Large \frac{\frac{1}{x^2}*2x}{2x}\]

Answer 4

Any questions on that?

Answer 5

one i understand what you did

Answer 6

what smoothmath said
then to a tiny big of algebra,
you will see that the limit is not 1

Answer 7

none*

Answer 8

Isn't it though, Satellite?

Answer 9

Cool, if you understand those steps, then all you have to do is evaluate at x=1, which gives you 2/2 =1

Answer 10

So yeah, you had the right answer.

Answer 11

yes all good :) thanks

Answer 12

Unless I made some mistake or Satellite is seeing something I don't.

Answer 13

I dont know but that would be 2 to 1

Answer 14

My pleasure! =D

Answer 15

oh yes, it is one
for an idiotic moment i was replacing x by 0!