How to prove limit n->infinity (n^log n) /(n^sqrt n) = 0

Since both numerator and denominator tends to infinity, I tried applying L'Hopitals rule but that complicated things. I'm stuck, how should I proceed? I'm bad at calculus.

Note that
lim(n→∞) n^(log n) / n^(√n)
= lim(n→∞) n^(log n - √n).
= lim(n→∞) e^[ln (n^(log n - √n))].

Next, observe that since
lim(n→∞) ln n/√n = lim(n→∞) (1/n)/(1/(2√n)) = lim(n→∞) 2/√n = 0,

lim(n→∞) (ln n - √n)
= lim(n→∞) √n * (ln n/√n - 1)
= -∞.
--------------
So, lim(n→∞) ln (n^(log n - √n))
= lim(n→∞) (ln n - √n) ln n
= -∞, by the previous observation.

Therefore,
lim(n→∞) n^(log n) / n^(√n)
= lim(n→∞) e^[ln (n^(log n - √n))]
= 0.

I hope this helps!

This was a tricky limit question; it was a matter of presenting this in the right order.

Report Abuse