What does $f(x)=\dfrac{e^{(4x)\pi\cdot i}}{x}$ do as $x\to0$ from the right?

Question

What does $f:\Bbb R\to\Bbb C:: f(x)=\dfrac{e^{(4x)\pi\cdot i}}{x}$ do as $x\to0$ from the right?

This is a spiral parametrised by $x$. I can see the real part goes to infinity and it does so in the upper-right quadrant, but is there a limit to the slope or height of the spiral in terms of the change in the real versus the imaginary part as it approaches infinity?

Differentiating I get $\frac{df}{dx}= \dfrac{i e^{4 i\pi x} (4 π x + i)}{x^2}$ and $\Im(\frac{df}{dx})=0$ at $x=0$ suggesting the imaginary part may have an upper bound.

Using a computer I get $\Im\left(\lim_{ x \to 0} \left(\dfrac{e^{(4x)\pi\cdot i}}{x}\right)\right)=0$

But then I get $\lim_{ x \to 0} \Im\left(\left(\dfrac{e^{(4x)\pi\cdot i}}{x}\right)\right)=4\pi$ so I'm a bit stuck what to think.

Answer 1

We have that

$$f(x)=\dfrac{e^{(4x)\pi\cdot i}}{x}=\dfrac{\cos (4\pi x)+i\sin(4\pi x)}{x}.$$

So $$\lim_{x\to 0^+}\dfrac{\mathcal{R}(f(x))}{x}=+\infty$$ and

$$\lim_{x\to 0^+}\dfrac{\mathcal{I}(f(x))}{x}=4\pi.$$

Answer 2

The real part is $$\mathrm{Re}(f(x)) = \frac{\cos (4\pi x)}{x}$$

which, as you observed, tends to $+\infty$ as $x$ goes to $0$ from the right:

$$\lim_{x\to 0^{+}} \frac{\cos (4\pi x)}{x} = \infty$$

Note that the same expression tends to $-\infty$ as $x$ approaches $0$ from the left.

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The imaginary part is

$$\mathrm{Im}(f(x))= \frac{\sin (4\pi x)}{x}$$

In this case, the two-sided limit at $x=0$ does exist:

$$\lim_{x\to 0} \frac{\sin (4\pi x)}{x} = 4\pi$$