lim x ^ ln x / ln x ^ x ? x>+infinitely?

Answer 1

lim x →∞ [x ^ ln x / ln x ^ x]
= lim x →∞ [x^ ln x / xln x]
= lim x →∞ [x^ (ln x - 1) / ln x]
= lim x →∞ [e^{ln x * (ln x - 1)} / ln x]
= lim x →∞ [{1/x(ln x - 1) + ln x*1/x}*e^{ln x * (ln x - 1)}/1/x] by l'Hopital's Rule
= lim x →∞ [(2ln x - 1) * e^{ln x * (ln x - 1)}
= lim x →∞ [(2ln x - 1) * x^ln (x/e)]
= ∞

Please ignore as according to Excel this is clearly incorrect!!
The limit = 0

My error arose from interpreting lnx ^ x as ln(x^x) rather than (ln x) ^ x

Let x = e^p. Then lnx = p
lim x →∞ [x ^ ln x / ln x ^ x] Let x = e^p Then lnx = p
= lim p →∞ [e^p²/p^(e^p)] and as x→∞, p→∞
= lim p →∞ [e^p²/e^(lnp*e^p)]
= lim p →∞ [e^(p² - lnp.e^p)]
Examining the index p² - lnp.e^p
Lim p →∞ (p² - lnp.e^p)
= Lim p →∞ (2p - 1/p.e^p - lnp.e^p)
= -∞
So lim p →∞ [e^(p² - lnp.e^p)] = 0

Answer 2

Let your limes be L. It is somewhat easier if you replace y = ln x. Then your expression is: L = e^(y^2)/y^(e^y) for y > infinity. L'hospital rule gives L = 2y*e^(y^2)/(lny*y^(e^y) + e^y * y^(e^y - 1)) which leads to L = L * (2y^2/(e^y + ylny)) which leads to L = L * 0. The only solution of this equation is L = 0.

Answer 3

=infinity

Answer 4

erm, ok then?

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