# For what rays does the limit $\lim\limits_{z\to\infty} e^x$ exist?

The original question was: "For what rays (beginning at the origin) does $\lim\limits_{z\to \infty} |e^z|$ exist?" I have simplified $|e^z|$ to equal $e^x$, though.

From here, I am not sure how to find the limit. I have a feeling it is either when $x$ is strictly $0$ or when $x$ is strictly finite. But I can't figure out how I would show that because it is when ${z\to \infty}$ and not when ${x\rightarrow \infty}$.

Obviously, we know that $\lim\limits_{x\to \infty} e^x = \infty$. But this is a different question.

Any help is appreciated. Thanks a lot!

NOTE: I accidently had $\lim\limits_{x\to \infty} |e^z|$ before, but it should be $\lim\limits_{z\to \infty} |e^z|$, as I have now edited the problem. Thanks and sorry about the confusion.

• I assume you are taking $z=x+iy$ when you talk about a limit in $z$ of an equation in $x$? Commented Nov 25, 2013 at 13:53
• To answer @abiessu's comment, yes. In this scenario, $z=x+iy$. Commented Nov 25, 2013 at 13:58

Note that if $z = x+iy$, then $|e^z| = |e^x(\cos y +i\sin y)| = |e^x| = e^x$, since $\cos y +i\sin y$ lies on the unit circle in $\mathbb C$.

There seems to be some confusion in the question as to whether we are working on all rays in the complex plane (starting from $0$), or only those for which we have $x \to \infty$. The answer below covers all cases, but if you only want the later, then you can ignore the bits of the answer you do not need.

Thus we need only consider the behaviour of $e^x$ as $x$ changes as we go along a ray.

Let $\theta$ be the angle the ray makes with the positive $x$-axis.

If $-\pi/2 < \theta < \pi/2$, then $x$ increases (towards infinity) as we move along the ray away from the origin, so $|e^z|$ tends to infinity.

If $\theta = \pm \pi/2$, then $x$ is always zero along the ray, so the limit is $e^0=1$.

for other rays (pointing to the left from the origin) we have $x \to -\infty$ , so that $|f(z)| \to 0$.

• I don't think it is defined along $\theta = \pm \frac \pi 2$, as you can't have $x \to \infty$ along those angles. Commented Nov 25, 2013 at 14:20
• @RossMillikan Yes - there seems t be some confusion between the question in the title and the question in the body! Commented Nov 25, 2013 at 14:21