I'll give it a try...

Posted by Joel on August 04, 2002 at 23:13:17:

In Reply to: If anyone's warmed up, please demonstrate a valid method for this limit problem's answer: posted by Dr. Andrew Coultos on August 02, 2002 at 14:06:08:

: In terms of e, please find the exact value of

: lim(x->0+){(x + x^x)^[1/(1 - x^x)]}.

= lim(x->0+) e ^ {ln(x + x^x)^[1/(1 - x^x)]}
= lim(x->0+) e ^ {[1/(1 - x^x)]* ln(x + x^x)}

Here, I'm not certain, but I think I can do this:
= e ^{lim(x->0+) [1/(1 - x^x)]* ln(x + x^x)}
If so, then I just work on the limit lim(x->0+) {[1/(1 - x^x)]* ln(x + x^x)}
= lim(x->0+) {ln(x + x^x)/(1 - x^x)} which is an indeterminate form of type 0/0 so can be solved using L'Hospital'x Rule.

lim(x->0+){d/dx [ln(x + x^x)] / d/dx [1 - x^x]}
= lim(x->0+){ [ 1/(x + x^x) ] * [ 1 + (x^x)*(1 + ln(x)) ] / [ -(x^x)*(1 + ln(x)) ] }
= lim(x->0+){ [ 1/(x + x^x) ] + [ (x^x)*(1 + ln(x))/(x + x^x) ] / [ -(x^x)*(1 + ln(x)) ] }
= lim(x->0+){ - [ 1 + (x^x)*(1 + ln(x)) ] / [ (x + x^x) * (x^x) * (1 + ln(x)) ] }
= lim(x->0+) { -1/[ (x + x^x) * (x^x) * (1 + ln(x))] - 1/(x + x^x) }
as x->0+, x^x -> 1 and ln(x) -> -oo so the first term, { -1/[ (x + x^x) * (x^x) * (1 + ln(x))] } -> 0
and the second term, { -1 / (x + x^x) } -> -1

So, the lim(x->0+) {ln(x + x^x)/(1 - x^x)} = -1 and therefore,

lim(x->0+){(x + x^x)^[1/(1 - x^x)]} = e^(-1)

How's that?

Follow Ups:

• Joel, I agree with your limit's value. [No message follows.] Dr. Andrew Coultos 10:23:07 08/06/02 (1)
  • Re: Joel, I agree with your limit's value. [No message follows.] sniper 19:34:29 08/15/02 (0)

Post a Followup