Question

Find the limit of ln (1-x^2)/lnx when x tends to infinity

Answer 1

undefined ?

Answer 2

@sauravshakya thats the wrong identity ..

Answer 3

oops

Answer 4

Ya..... thats wrong

Answer 5

well log(-ve) is undefined,,so wont the limit, simply not exist..?

Answer 6

u know L'hopital's rule ? @duwduuw

Answer 7

2!

Answer 8

Ya.... @hartnn has correct approach

Answer 9

(not factorial, merely excitement)

Answer 10

I think its 0

Answer 11

\[\frac{ \log(1-x )+\log_{}(1+x ) }{ logx }\]
evaluate seperately

Answer 12

i was thinking exactly that jonask

Answer 13

Yep i do know the hospitals rule with the derivatives!

Answer 14

are they difficult graphs?

Answer 15

so did u try differentiating numerator and denominator separately ?

Answer 16

"the hospitals rule"
^.^

Answer 17

Again got the same.... its 0

Answer 18

\[\frac{ \infty+\infty }{ \infty }=2\]
is there such thing

Answer 19

did u get (2x^2)/(1-x^2) ??
and answer = 2

Answer 20

@Jonask there is no such thing

Answer 21

log(1-x^2) at x-->infinity = log(-infinity) -->undefined,,i'd simply stick to my ans that it is undefined..

Answer 22

yup

Answer 23

One week ago, every time a limit showed up, everyone immediately screamed "l'Hopital's !". Now it seems we work up to it slowly.

Answer 24

@shubhamsrg is correct

Answer 25

wolf says its 2 !!!

Answer 26

@shubhamsrg is correct

Answer 27

You're all fired. Except @hartnn . He's promoted to assistant to the regional manager.

Answer 28

please stop joking @Algebraic!

Answer 29

graphycally

Answer 30

use l'Hopital's
/thread

Answer 31

@duwduuw u differentiated numerator and denominator, right ? what did u get ?

Answer 32

you sure the question doesnt have ln(x^2 -1) there??
had it been that,,surely ans is 2 then..
otherwise ..you all know my answer! :P

Answer 33

2x/xln(1+x^2)

Answer 34

\[ \ln (1 + x) = \ln x + \ln (1/x + 1) \\
\ln(1 - x) = \ln(-x) + \ln (-1/x + 1)\]

Answer 35

interesting identity
http://www.wolframalpha.com/input/?i=limit+x-%3Einf+ln%28-x%29%2Fln%28x%29
L'hopital is much simpler.

Answer 36

ohh!! :O

Answer 37

ln term will not come in differentiating num and denom separately.
u get (2x^2)/(1-x^2)

Answer 38

even there L'Hopital's is used !

Answer 39

since log(-1) = i(pi) ,,,its some constant,,but still can we ignore it in front of infinity ?

Answer 40

you can do that without L'hopital's rule ...
you need to make a slight modification of this identity
http://www.wolframalpha.com/input/?i=limit+x-%3E0+%28-x%29^x

Answer 41

aha.. @mukushla will you confirm this please ?

Answer 42

Oooooo nevermind, i meant to say 2x^2/1+x^2

Answer 43

I agree hartnn

Answer 44

u get 2x^2/(1-x^2) = 2/(1/x^2-1)
and when u put x=infinity, u get 2.

Answer 45

Exactly, so 2 is the limit.
+1 for hartnn haha

Answer 46

thank u :)

Answer 47

What am I? Chopped liver?

Answer 48

daw

Answer 49

Haha sorry Just saw that you had the answer firstttt