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

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

Mathematics
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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
The limit is as r approaches 0+, but the answer is supposed to be 0 and I just can't prove it...
This is an indeterminate of the form 0^(0) so try using L'Hopitals Rule

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my mistake its actually (-infinity)^(0) but still and indeterminate, so L'Hopitals should work
the answer is 0
does anyone want the method
yes please!
can nobody please explain to me how to apply de l'hopital to this? or any other method?
hello friend, it was my mistake. I thought it was r^2/ln(r^2)?
else the answer is infinity. I can explain you the L'hospital rule
If you get 0^0 form or 0/0 then you can differentiate the given function which is r^2*ln(r^2) here
when u find d/dr(r^2*ln(r^2) the answer is 2rln(r^2)+2r
this is again of the form 0^0 therefore u differentiate it again
so u wud get 2ln(r^2)+6 which is again log0=infinity, when r tends to 0
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
therefore u wud get an infinity
okay
but wouldn't the limit of 2ln(r^2)+ 6 be -infinity?
so u can again differentiate
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
yes but 50% of the time its -inf and the other 50% its +inf.. so which one is the right answer?
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
so ur answer is DNE
Thanks Dipin!
ur welcome

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