How do I evaluate this limit ? that's the only thing I know about a.. any ideas ?

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answers:
kb say: First of all, note that
lim(x→∞) x^a (x - ln(e^x + 1))
= lim(x→∞) x^a [x - ln(e^x (1 + 1/e^x))]
= lim(x→∞) x^a [x - (ln(e^x) + ln(1 + e^(-x)))], product rule for logs
= lim(x→∞) x^a [x - (x + ln(1 + e^(-x)))]
= lim(x→∞) -x^(a+1) ln(1 + e^(-x))).

Case 1: If a+1 < 0, then the limit equals 0 * 0 = 0.
Case 2: If a+1 = 0, then the limit equals 1 * 0 = 0.

Case 3: If a+1 > 0, rewrite the limit as follows:
lim(x→∞) -ln(1 + e^(-x)) / x^(-(a+1)); now of the form 0/0
= lim(x→∞) [e^(-x)/(1 + e^(-x))] / [-(a+1) x^(-(a+2))], by L'Hopital's Rule
= (-1/(a+1)) * lim(x→∞) x^(a+2) * [e^(-x)/(1 + e^(-x))]
= (-1/(a+1)) * lim(x→∞) x^(a+2) * [1/(e^x + 1)], multiplying each term in the fraction by e^x
= (-1/(a+1)) * lim(x→∞) x^(a+2)/(e^x + 1)
= (-1/(a+1)) * lim(x→∞) (a+2)x^(a+1)/e^x, by L'Hopital's Rule (∞/∞)
= (-(a+2)/(a+1)) * lim(x→∞) x^(a+1)/e^x
= (-(a+2)/(a+1)) * 0, since a+1 > 0, and e^x always grows faster than any positive power of x
= 0.
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Hence, lim(x→∞) x^a (x - ln(e^x + 1)) for all real 'a'.

I hope this helps!
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Indikos say: I first prove f(x) = x - ln(1+x) is increasing for x > 0
f'(x) = 1 - 1/(x+1)
And for x > 0 , f'(x) > 0 and therefore f(x) is increasing for x > 1
And we have x > ln(1+x)

Consider g(x) = x^a ( x - ln(e^x + 1))
= x^a ( x - ln(e^x ( 1 + e^-x))
= x^a ( x - ln(e^x) - ln( 1 + e^-x))
= x^a ( x - x - ln(1 + e^-x) )
= -x^a * ln(1 + e^-x)

Now x^a * ln(1 + e^-x)
<= x^a * e^-x .... since e^-x - ln(1+e^-x) = 0 for all x

If a <= 0 then x^a <= 1 for all x > 1
and lim x -> oo x^a * e^-x <= lim x -> oo e^-x = 0

If a > 0 then let n = [a] + 1 where [] is the greatest integer function
and we have 0 <= x^a * e^-x <= x^n e^-x
Also x^n e^-x = x^n /( 1 + x + ........ x^n/n! + x^(n+1)/(n+1)! + ....... )
<= x^n /( x^(n+1)/(n+1)! + ....... )
< = 1/(x/(n+1)! + ........... )

and as lim x -> oo 1/( x/(n+1)! = 0
Therefore lim x -> oo x^a * e^-x = 0 for all a
and the desired result follows.
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Puzzling say: ln(e^x + 1) is very close to ln(e^x) for very large values of x. And hence that is close to x.

So (x - ln(e^x + 1)) approaches 0 and even when it is multiplied by x^a still goes to 0, just less quickly.
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