Limit of $n^{A\lg n}\cdot 2^{n-n^A}$ for $A > 1$

Question

Can anybody think of a straightforward way to see that this limit equals $0$ for all $A > 1$? I can only think of applying L'Hopital's rule, which leads to a godawful mess. Perhaps there's a useful upper bound for $n^{\lg n}$?

Answer 1

$$n^{A\lg n}=2^{A\ln^2n}$$

Thus, we may rewrite it as

$$2^{A\lg^2n+n-n^A}$$

As $n\to\infty$ the $-n^A$ dominates, since $A\lg^2n=\mathcal O(n)$ and $n^A>n^1$. Thus, as $n\to\infty$, the limit is simply $0$.