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

*To*: mathgroup at smc.vnet.net*Subject*: [mg8335] Re: [mg8215] L'Hopital's rule contradiction --- what am I messing up?*From*: seanross at worldnet.att.net*Date*: Sun, 24 Aug 1997 13:24:51 -0400*Sender*: owner-wri-mathgroup at wolfram.com

Tom Chwastyk wrote: > > 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? > > Thanks in advance, > > 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 I think you are mis-applying L'Hopitals rule. It states that if the limit of x/y is indeterminate, then x'/y' has the same limit. Your first case was p Log p and you apply LH to get 1 * 1/p. Instead you needed to write either Log p /(1/p) in which case both numerator and denominator are infinity or p/(1/Log p) in which case they are zero. Apply LH to the correct numerator and denominator and I bet you will get a better answer. The second example you give is a case of cancelling errors.

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

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