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Rudin 8.5a

  • Thread starter ehrenfest
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  • #1
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[SOLVED] rudin 8.5a

Homework Statement


Find the following limit:

[tex]\lim_{x \to 0} \frac{e-(1+x)^{1/x}}{x}[/tex]


Homework Equations


[tex]e = \lim_{x\to 0} (1+x)^{1/x}[/tex]

[tex]a^b = e^{b \log a}[/tex]


The Attempt at a Solution


Is this the derivative of something? I doubt it is but that is the only way I would know how to do this.
 

Answers and Replies

  • #2
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anyone?
 
  • #3
MathematicalPhysicist
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Well, you could use l'hopital rule here, but I guess you havent learnt it yet.
besides this I don't see a handy trick here.
 
  • #4
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Well, you could use l'hopital rule here, but I guess you havent learnt it yet.
besides this I don't see a handy trick here.
Of course I know l'Hopital. Using it here requires me to calculate:

[tex]\lim_{x \to 0} \left(\frac{1}{x+x^2}-\frac{\log (1+x)}{x^2} \right) \exp \frac{\log (1+x)}{x}[/tex]

which I have no idea how to do. Of course I could probably use l'Hopital again but then it would just get messier. Of course I only need to find the limit of the expression in parenthesis but I am not really even sure if that exists.
 
Last edited:
  • #5
lurflurf
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first show
e/(2/x+2)<e-(1+x)^(1/x)<e/(2/x+1)
then squeeze
 
  • #6
MathematicalPhysicist
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the derivative of the numerator is:
(log(1+x)/x)'(-(1+x)^(1/x))=
(-(1+x)^(1/x))*((x/(1+x))-log(x+1))/x^2
now you should be taking l'hopital on :((x/(1+x))-log(x+1))/x^2
you can check if you got the right answer through mathematica.

p.s
you shouldn't pay much attention to those cacluations' problems, pay more attention to theoretical questions in rudin's book.
 
  • #7
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now you should be taking l'hopital on :((x/(1+x))-log(x+1))/x^2
Yay, I got it:

[tex]\lim_{x \to 0} \frac{(x/(1+x))-\log(x+1)}{x^2} = \lim_{x \to 0} \frac{1/(1+x)^2-1/(1+x)}{2x} = \lim_{x \to 0} \frac{-2/(1+x)^3+1/(1+x)^2}{2} = -1/2[/tex]

So the answer is e/2. Apparently lurflurf was right but I have no idea how he got those inequalities.


I think these computational problems are good preparation for the Putnam.
 

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