Let $f(n) = 2^\sqrt{n}$ and $g \in \mathbb{R}[T] \implies g = \mathcal{O}(f)$.
My attempt: $$\lim_{n\to\infty}\frac{g}{f(n)} = \lim_{n\to\infty}\frac{\sum_{k=0}^da_iT^i}{2^{\sqrt{n}}} = \lim_{n\to\infty}\sum_{k=0}^da_i\frac{T^i}{2^{\sqrt{n}}} = \sum_{k=1}^d \left(\lim_{n\to\infty}\frac{a_iT^i}{2^{\sqrt{n}}}\right) = 0$$
Therefore
$$\lim_{n\to\infty}\frac{g}{f(n)} = 0 \implies g \leq f(n)\implies g = \mathcal{O}(f),$$
But I'm not sure if this is correct
Where can I find more exercises of this type?