How does the lim(a-0+) (-a^2(2lna-1))/2=0?

Question

How does the lim(a->0+) (-a^2(2lna-1))/2=0?

anonymous · Answer

The a^2 part goes to zero faster than the ln(a) part goes to negative infinity.

anonymous · Answer

Is this something that I should be able to prove? Or do you think that I could just state that

anonymous · Answer

You can prove it using L'hospital's rule if you'd like.

anonymous · Answer

Ok. Can you show me how to do that?

anonymous · Answer

Because I feel like that would make it a mess

anonymous · Answer

Apologies, having a bit of trouble with my browser at the moment... you would write it as

ln(a) / (1/a^2)

Then taking the derivative of the top and bottom, you get

1/a / (-2/a^3)

which can also be written as 

-a^2/2

Which goes to zero as a->0.

anonymous · Answer

No worries. I see. Thank you!!!

anonymous · Answer

No problem.  Though you should always show your work when necessary, it is occasionally useful to remember than ln(x) is overpowered by the behavior of any power of x, so x*ln(x) -> 0 as x->0 and ln(x)/x -> 0 as x->infinity.