$i^{1/n}$ when $n \to \infty$

As noticed and discussed in detail by davidlowryduda and Lubin, the existence of the limit depends upon the definition we assume for $i^\frac1n=\sqrt[n] i$.

Notably, according to the standard definition, we have

$$i=e^{i\frac{\pi}2+2ik\pi}\implies \sqrt[n] i=e^{i\frac{\pi}{2n}+i\frac{2k\pi}n} \quad k=0,1,\ldots,n-1$$

which is a multivalued function and therefore we can't define/determine the existence or a value for the limit without defining a "rule" to choose one of the $n$ root.

For example if we assign to $\sqrt[n] i$ the root corresponding to a fixed value for $k=\bar k$ of course the limit is $1$, indeed

$$e^{i\frac{\pi}{2n}+i\frac{2\bar k\pi}n} \to e^0=1$$

but if we choose a value of $k$ depending upon $n$ the limit might be different and assume any value $e^{i\theta}$ for complex number on the unit circle.

The set of $n$th roots of $i$ become equidistributed around the unit circle as $n \to \infty$, and so on the face of it this is a delicate question.

It seems to me that the entire crux of this matter comes down to defining what you mean by $i^{1/n}$. You see, this isn't single-valued for complex numbers. Indeed, for any fixed $n$, we might define $$a^{1/n} = e^{\frac{1}{n}\log a} = e^{\frac{1}{n}(\log \lvert a \rvert + 2\pi i k \arg(a))}$$ for any value of $k$ (although in practice, it's sufficient to choose $0 \leq k < n$). A priori, for each $n$th root, one can take a different $k$ and use that branch of the logarithm.

Some of the other answers seem to be assuming that you are defining the $n$th root of $i$ in a way that fixes $k$, and takes that sequence of branches of the log. But this is neither obvious nor necessarily true. If you fix $k$, then the limit is $1$. If you choose to let $k \approx n/2$ as $n \to \infty$, then the limit will be $-1$. One might say that this is an odd choice of $k$, but I think it highlights that the choice of $k$ (or rather, exactly how one defines the $n$th root) matters significantly.

Write

$$i=e^{\frac\pi2i+2k\pi i}\implies i^\frac1n=e^{\frac\pi{2n}i+\frac{2k\pi i}n}\xrightarrow[n\to\infty]{}e^0=1$$

So even the number's argument is not uniquely defined, the limit still is one (here we assume the exponential function's continuous on the complex plane. Not so big an assumption...)