Limit as x ---> infinity of x^(1/(1+2lnx))

im stumped...

L = lim x^(1/(1 + 2lnx))
......x-> ∞
L = lim ln(x^(1/(1 + 2lnx)))
......x-> ∞
L = lim 1/(1 + 2lnx) lnx
......x-> ∞
lnL = lim lnx/(1 + 2lnx)...use L'Hopital's rule next
........x-> ∞
lnL = lim (1/x)/(2/x)
........x-> ∞
lnL = 1/2
L = e^(1/2)

Look instead at the limit as x -> infinity of ln(x^(1/(1+2ln(x)))) = (1/(1+2ln(x)))*ln(x).
Thus we have lim(x->inf.)(ln(x) / (1 + 2ln(x))) which is in the form infinity/infinity
so we can use L'Hopital's Rule to equate this to lim(x->inf.)((1/x) / (2/x))) = 1/2.
Thus since the natural log of the expression goes to 1/2 as x goes to infinity,
the expression itself will go to e^(1/2) as x goes to infinity.

lim x^(1 / (1 + 2ln x))
x->∞
direct substitution yields the indeterminate form ∞^0; so you must try another technique
y = lim (x -> ∞) x^(1 / (1 + 2ln x))
ln y = lim (x -> ∞) ln [x^(1 / (1 + 2ln x))]
ln y = lim (x -> ∞) (1 / (1 + 2ln x)) ln x
ln y = lim (x -> ∞) ln x / (1 + 2ln x) ---> use l'Hôspital's rule
ln y = lim (x -> ∞) d/dx[ln x] / d/dx[(1 + 2ln x)]
ln y = lim (x -> ∞) (1 / x) / (2 / x)
ln y = lim (x -> ∞) 1/2
ln y = 1/2
y = √e

lim x^(1 / (1 + 2ln x)) = √e
x->∞

Hope that helps!