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MathGroup Archive 1997

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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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