       L'Hopital's rule contradiction --- what am I messing up?

• To: mathgroup at smc.vnet.net
• Subject: [mg8215] L'Hopital's rule contradiction --- what am I messing up?
• From: "Tom Chwastyk" <chwastyk at nrl.navy.mil>
• Date: Tue, 19 Aug 1997 21:11:53 -0400
• Organization: US Naval Research Laboratory
• Sender: owner-wri-mathgroup at wolfram.com

```Yesterday I was writing something dealing with the entropy of a probability
distribution s(p)=sum_i p_i log p_i. I wondered if I had to exclude p_i=0
from the sum, so I looked at

limit_(p->0) p log p = 0 (-infinity) .

This is an indeterminate form, so I tried applying L'Hopital's rule
(limit of an indeterminate form two-factor product is the limit of the
product of the derivatives of the factors) and got

.... = limit_(p->0) (1) (1/p) = infinity .

This seemed odd (I didn't see the exclusion all over the literature), so I
asked Mathematica for Limit[p Log[p], p->0] and got back 0. I am confident
that 0 is correct. For one thing, I looked at plots of p and log p; p seems
to go to 0 faster than log p goes to -infinity. Second, I found the MMa
book even includes (with evident pride) the example Limit[x Log[x],x->0]
giving 0 'even though there's not a power series for Log[x] about 0'.
Third, I looked at q=log p, p=e^q with the equivalent

limit_(q->-infinity) e^q q = 0 (-infinity);

this is indeterminate again, but now L'Hopital gives

.... = limit(q->-infinity) (e^q) (1) = (0) (1) = 0.

Question: what am I messing up or overlooking in the first version?

Tom
--
Tom Chwastyk ["Fosdick". Polish: CH='H, W=F (here),
A=AH, Y=I. Soften strict 'HFAH-stik.... :-) ]
Naval Research Laboratory Code 6383
Wash. DC 20375-5343    (202) 767 2567

```

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