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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
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