Does the limit of $e^{-1/z^4}$ as $z\to 0$ exist?

The question is as follows (note that $z = x + yi$):

Does $\lim_\limits{z \to 0} e^{-1/z^4}$ exist?

I’ve shown that

$\lim_\limits{x \to 0} e^{-1/x^4} = 0$

is equal to

$\lim_\limits{y \to 0} e^{-1/(iy)^4} = 0$

To prove both parts, but I’m not sure this is how it is done. I’ve been away from class for surgery, so all I have to work off of is what I’ve read online and in the textbook.

The problem is, it seems that with problems such as this one:

$\lim_\limits{z \to 2} \frac{z^2 + 3}{iz} = \frac{-7i}{2}$

All you have to do is substitute in 2 directly. I can’t seem to figure out how to approach these problems, so hopefully all I have explained makes sense. Thanks for any explanations!

Solutions Collecting From Web of "Does the limit of $e^{-1/z^4}$ as $z\to 0$ exist?"

Limits in complex plane are not really different from limits in multivariable calculus. (You can find plenty of examples in my catalog of multivariable limits). The strategy for considering the limit as $z\to a$:

  1. Check if you already know that the function is continuous at $a$ (e.g., it’s a rational function with nonzero denominator at $a$). If this is so, then the limit is $f(a)$.

  2. Try some lines of approach. If some of them give different limit values, the limit does not exist.

  3. If all lines appear to give the same limit value, try some squeeze theorem argument; this step depends very much on what your function is.

  4. If all of the above fails; set the problem aside for a while and try it again later, possibly also considering curves in addition to lines in 2.

In your example $\exp(-1/z^4)$, the horizontal and vertical directions of approach yield $0$. But another line, such as $y=x$, yields $\infty$ because $(x+ix)^4 = (1+i)^4x^4 = -4 x^4$, leading to $\exp(1/(4x^4))$. So the limit at $0$ does not exist.