We say that that a function $f$ is meromorphic at infinity if the function $g(z) = f(1/z)$ has a pole at $0$.
Suppose that an entire function $f$ is holomorphic at infinity. Show that $f$ is constant.
My attempt:
$f(1/z) = z^{-n}×h(z)$ and so $f'(1/z) = -nz^{-n-1}×h(z) + z^{-n}×h'(z)$, and so we can write $f'(1/z) = \frac{-n}{z^{n+1}}×h(z) + \frac{h'(z)}{z^n}$, and so taking the limit as $n \to$ infinity, we get $f'(1/z) = 0$ so $f$ is constant
Is my attempt correct?
An entire function meromorphic at infinity is not necessarily constant: as an example, take a non-constant polinomial $P(z)$.
Your attempt is flawed because
$$f'\left(\frac{1}{z}\right)=\frac{h'(z)z-nh(z)}{z^{n+1}}$$
does not imply that $f'(1/z)\to 0$, as$h(z)$ may grow faster then $z^n$ (take $f(z)=z$ to see it explicitly).
It is true, however, that every entire function holomorphic infinity is constant. To prove this, note that $f(1/z)$ is bounded in the unit disc and thus $f(z)$ is bounded for all $|z|>1$. By Liouville's theorem, $f$ is constant.
As a final note, there is a possible characterization of entire functions meromorphic at infinity: they are exactly the polynomials.