what's the limit as n--> infinity of 2(n!^1/n)?

Answer 1

∞

Answer 2

By Stirling's formula (see link), n!^(1/n) will be asymptotic to
√(2πn)^(1/n) . (n/e)
= (2πn)^(1/(2n)) . n/e
Now lim (n-> ∞) (2πn)^(1/(2n)) = 1*, so 2(n!^(1/n)) will be asymptotic to 2 (1) . n/e and hence we will have
lim (n->∞) 2(n!^(1/n)) = ∞.

* This should be obvious, but if not, put n = e^k and note that 2π < e^2, so (2πn)^(1/(2n)) < (e^(k+2))^(1/(2e^k))
= e^((k+2)/(2e^k)) -> e^0 = 1.