I am looking for a good book on limits of functions and clever ways of calculating them. Any ideas? some kind of handbook.For example, how can I calculate the limit of (1 - y^2) * y * log((1+y)(/1-y)) * log((1+y)(/1-y)) as y -> 1?? I could work out the series expansion and plug in y = 1 but I suppose there must be a better way!

If I got the limit you want to find right then seems to me that Lopital's rule is enoughnow, so the limit is 0.

This case is straightforward because you have a polynomial factor, 1- y, tending to zero, multiplying a (power of a) Log factor that's tending to infinity: the polynomial will always win. (Complete the square to get 1-y !)The general idea is to compare the rates at which the factors are heading towards 0 or infty; i.e. use L'Hopital's Rule. Factors heading for any other values can simply be substituted by those values.Inserting the finite values and then substituting x = 1 - y gives L= 2 lim_{x->0} x Log^2 (2/x).Now substituting z = 1/x givesL = 2 lim_{z->infty} (1/z) Log^2 (z), which after 2 applications of L'Hopital's Rule gives L = 0.

I knew I misspelled L'Hopital

Hi,I guess my answer might sound provocative, but what about mathematica?Input : Limit[(1 - y^2)*y*Log[(1 + y)/(1 - y)]*Log[(1 + y)/(1 - y)], y -> 1]Output : 0This is definitely not a methodological answer, but you'll have your answer in many cases, which will give you a hint of what to look for if you want to be able to find it on your own (pen and paper).blondie.PS : Guillaume François Antoine, marquis de L'Hôpital (1661 - 1704).In old french, 'ô' was 'os' so it is sometimes L'Hospital's rule. It seems L'Hospital was giving money to Jean Bernoulli (see wikipedia in french),so that the rule should be Bernoulli's rule.Anyway, that does not help, does it?

Problem is there's no way of telling -- without analysis -- if the computer output is the result of a bug. Try this in Mathematica 5.0:In[1]:= HarmonicNumber[Infinity, 3]Out[1]= Zeta[3]In[2]:= Limit[HarmonicNumber[n, 3], n -> Infinity]Out[2]= InfinityJust like magic!

Bugs are always a possibility, but for this one, a Plot[...] plus a couple terms of Series[...]give one great confidence in Mathematica's Limit[...] calculation.

Thanks a lot guys for the feedback. The value "0" is what I had needed. Using Mathematica seems like a good idea, as Blondie has suggested For the other case(1-y^2) * log((1+y)/(1-y)) I used a series expansion and the root test at y = 1. So now I can proceed.//At some stage I will need to look at (1-y^2)^a * log^b ((1+y)/(1-y)) with a = 1, 2 and b arbitrary real number, possibly even negative.

If b <0 then it's easy because you have L = 0/infty = 0. If b=0 then L = 0 obviously.If b > 0, let c = smallest integer larger than b. Then apply my previous recipe, gettingL = 2^(2a) lim_{z->infty} (1/z^a) Log^b (z).Apply L'Hopital's rule c times, getL = b! (1/a)^c 2^(2a) lim_{z->infty} (1/z^a) Log^(b-c) (z) = 0because this is simply 1/(infty * infty).

QuoteAlan: Bugs are always a possibility, but for this one, a Plot[...] plus a couple terms of Series[...]give one great confidence in Mathematica's Limit[...] calculation.I dunno, I've seen too many examples of functions whose graphs look optically identical yet give vastly different (and wrong) physics.

QuoteOriginally posted by: INFIDELProblem is there's no way of telling -- without analysis -- if the computer output is the result of a bug. Try this in Mathematica 5.0:In[1]:= HarmonicNumber[Infinity, 3]Out[1]= Zeta[3]In[2]:= Limit[HarmonicNumber[n, 3], n -> Infinity]Out[2]= InfinityJust like magic!In Mathematica 6.0.2.0 I get:In[1]:= Limit[HarmonicNumber[n, 3], n -> Infinity]Out[1]= Zeta[3]

I tried now:x == 1/2 * log ((1+y)/(1-y)) <==> y = tanhxThen:(1-y^2) * log^2((1+y)/(1-y)) = (1 - tanh^2 (x)) * x^2y -> 1 <==> x -> infinity and the desired term goes to zero because no monomial in x can be faster than exp(x) as x -> infinity.