Question

lim at 0 of r^2*ln(r^2)?

Answer 1

Is this lim as r approaches 0? All you need to do for this one is plug 0 into all r's.

r^2*ln(r^2)?
(0)^2*ln((0)^2)
= DNE

Answer 2

The limit is as r approaches 0+, but the answer is supposed to be 0 and I just can't prove it...

Answer 3

This is an indeterminate of the form 0^(0) so try using L'Hopitals Rule

Answer 4

my mistake its actually (-infinity)^(0) but still and indeterminate, so L'Hopitals should work

Answer 5

the answer is 0

Answer 6

does anyone want the method

Answer 7

yes please!

Answer 8

can nobody please explain to me how to apply de l'hopital to this? or any other method?

Answer 9

hello friend, it was my mistake. I thought it was r^2/ln(r^2)?

Answer 10

else the answer is infinity. I can explain you the L'hospital rule

Answer 11

If you get 0^0 form or 0/0 then you can differentiate the given function which is r^2*ln(r^2) here

Answer 12

when u find d/dr(r^2*ln(r^2) the answer is 2rln(r^2)+2r

Answer 13

this is again of the form 0^0 therefore u differentiate it again

Answer 14

so u wud get 2ln(r^2)+6 which is again log0=infinity, when r tends to 0

Answer 15

again u differentiate it, u wud get 4/r
when r tends to zero, 4/r wud tend to infinity as any number divided by 0 is infinity

Answer 16

therefore u wud get an infinity

Answer 17

okay

Answer 18

but wouldn't the limit of 2ln(r^2)+ 6 be -infinity?

Answer 19

so u can again differentiate

Answer 20

when u differentiate it again and again and again, u r getting only infinity.differentiating means breaking into smaller parts. But when we add this delta x and delta y again and again we are still getting infinity so the answer wud be infinity

Answer 21

yes but 50% of the time its -inf and the other 50% its +inf.. so which one is the right answer?

Answer 22

nope its not -infinity, when u differentiate for the fourth, fifth, sixth and so on u wud only get +infinity apart from the 3 differentiation

Answer 23

so ur answer is DNE

Answer 24

Thanks Dipin!

Answer 25

ur welcome