# Entire function $f$ such that $\lim\limits_{z\rightarrow \infty}f(z)=0$ and $f(0)=1$?

The question is this: Does there exist an entire function $f$ such that $\lim_{z\rightarrow \infty}f(z)=0$ and $f(0)=1$.

I immediately would point to $f(z)=e^{-z}$. It is entire and satisfies the above. However my professor gives this answer:

No. Since entire functions are bounded on bounded sets, the condition $\lim_{z\rightarrow \infty}f(z)=0$ implies that $f$ is bounded on the whole complex plane. Then by Liouville's Theorem, $f(z)=f(0)$ and $\lim_{z\rightarrow \infty}f(z)=1\neq 0.$

Who is correct? Maybe the question is wrong and it is supposed to have the limit for $\pm \infty$?

• In the complex setting $\infty$ refers fo "complex infinity," which is the point at infinity on the Riemann sphere. A limit to complex infinity requires more than a limit to real infinity: the limit has to be the same "in all directions," for example. May 25 '16 at 17:00
• To sum up @QiaochuYuan's point, limits when $z\to\infty$ must be understood as limits when $|z|\to\infty$. That is, the statement $$\lim_{z\to\infty}g(z)=\ell$$ actually means: $$\forall\epsilon\in(0,+\infty)\quad\exists R\in(0,+\infty)\quad\forall z\in\mathbb C\quad |z|>R\implies|g(z)-\ell|<\epsilon.$$
– Did
May 25 '16 at 17:45

The function $z \mapsto e^{-z}$ doesn't satisfy the conditions. Actually the limit $$\lim_{z\to \infty} e^{-z}$$ doesn't exist. (For exemple let $z$ go to infinity on $i\mathbb{R}$) Hence it is not a counterexample.
The proposition is correct, by Liouville's theorem the function must by $1$ but then the limit is not $0$ so there are no such functions.