What's the limit of $e^z$ as $z$ approaches infinity?
I am given that the answer is "There is no such limit."
Is this correct, and if so, am I correct to demonstrate this by showing that as $y$ tends to infinity along the $y$-axis, the magnitude of $e^z$ remains $1$, i.e. it doesn't have infinite magnitude, thus it cannot be tending to infinity? And does this mean $e^z$ has an essential singularity at infinity?
Best Answer
Use the fact: $f$ has a pole of order $k$ at $z=z_0$ if and only if $$ \lim_{z\to z_0}(z-z_0)^k f(z)=L(\ne 0). $$
Relpacing $z$ by $\frac 1 z$ and consider the function $f(z)=e^{\frac 1 z}$. For each $k=0, 1, 2, \ldots $, the limit $$ \lim_{z\to 0}z^ke^{\frac 1 z} $$ does not exist (it suffices to check when $z$ is real).
Thus, $f$ has no poles(including removeable singularity) at the origin, and has essential singularity at the origin. (or more simply, it can be checked by expanding $f$ as Laurent series-using taylor series formula).
Therefore, $e^z$ has an essential singularity at $\infty$.