Re: Re: Problems with Limit, Log, E

*To*: mathgroup at smc.vnet.net*Subject*: [mg9235] Re: [mg9203] Re: [mg9149] Problems with Limit, Log, E*From*: Allan Hayes <hay at haystack.demon.co.uk>*Date*: Fri, 24 Oct 1997 01:00:50 -0400*Sender*: owner-wri-mathgroup at wolfram.com

Sean: In considering the failure of Mathematica to find Limit[ x - Log[1 + E^x], x -> Infinity] you look at x((1-Log[1+Exp[x]])/x) and (1-Log[1-Exp[x]]/x) / (1/x) and reject them. However, if we look at (x(1-Log[1+Exp[x]])/x we see that L'Hospital applies (since |x| --> Infinity as x-->Infinity - see note on L'Hospital's rule below) So we look at D[x(1-(Log[1+Exp[x]])),x]/D[x,x] 1 - (E^x*x)/(1 + E^x) - Log[1 + E^x] It is clear that the limit of this as x-->Infinity is -Infinity; and Mathematica agrees: Limit[%, x->Infinity] -Infinity L'HOSPITAL'S RULE A fairly strong form of L'Hospital's rule is: For a,b,c in the extended real line [-Infinity,Infinity] Suppose that c is in [a,b] and functions f and g are differentiable on (a,b) less c; and that as x --> c through (a,b) less c; either f(x) -->0 and g(x) --> 0 or |g(x)| --> Infinity Then if f'(x)/g'(x) --> lim as x --> c through (a,b) less c; we have f(x)/g(x) --> lim as x --> c through (a,b) less c. There is a proof of this in Walter Rudin, "Principles of Mathematical Analysis", McGraw-Hill,1953, p 82. Allan Hayes hay at haystack.demon.co.uk http://www.haystack.demon.co.uk/training.html voice:+44 (0)116 2714198 fax: +44 (0)116 2718642 Leicester, UK