Limit of (e-(1+x)^{1/x})/x as x approaches 0

[ehrenfest]

[SOLVED] rudin 8.5a

Homework Statement

Find the following limit:

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

Homework Equations

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

a^b = e^{b \log a}

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.

 

[ehrenfest]

anyone?

 

[MathematicalPhysicist]

Well, you could use l'hopital rule here, but I guess you haven't learned it yet.
besides this I don't see a handy trick here.

 

[ehrenfest]

Well, you could use l'hopital rule here, but I guess you haven't learned 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:

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

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.

 

[lurflurf]

first show
e/(2/x+2)<e-(1+x)^(1/x)<e/(2/x+1)
then squeeze

 

[MathematicalPhysicist]

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.

 

[ehrenfest]

now you should be taking l'hopital on :((x/(1+x))-log(x+1))/x^2

Yay, I got it:

\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

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.