Posted by sniper on August 15, 2002 at 19:34:29:
In Reply to: Joel, I agree with your limit's value. [No message follows.] posted by Dr. Andrew Coultos on August 06, 2002 at 10:23:07:
: : : 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?
Wow! Dr. C agrees!
He must have earned a day pass!