$i^{1/n}$ when $n \to \infty$ What would happen when $n \to \infty$ for $i^{1/n}$. Would it still be an imaginary number or would it simply be $i^0=1$?
 A: 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.
A: 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.
A: 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...)
A: I want to respond to this provocative question as an algebraist, not an analyst. As many have noted, there are $n$ distinct roots of the equation $X^n=i$, equally spaced around the unit circle. And algebra refuses to distinguish among them: any $n$-th root of $i$ will be as good as any other.
If you’re willing to go along with me in this refusal, then you must agree that every complex number $a+bi$ on the unit circle, i.e. satisfying $a^2+b^2=1$, is a limit, as $n\to\infty$, of $n$-th roots of $i$, as long as those roots are  properly chosen.
For, let $a+bi$ be such a point on the unit circle. Among the $n$ numbers $\zeta$ satisfying $\zeta^n=i$, there is one, call it $\zeta_n$, that is at a distance of at most $\pi/n$ from $a+bi$. So if we ask for $\lim_n\zeta_n$, the value in answer is our $a+bi$.
( But at least there are limits. In my preferred environment, the $p$-adic, any sequence of $n$-th roots of $i$ will have no limit at all. )
A: You can write $$ i^{\frac{1}{n}}=e^{\frac{1}{n}\ln(i)}$$
Now, $$\lim\limits_{n\rightarrow\infty}i^{\frac{1}{n}}=\lim\limits_{n\rightarrow\infty}e^{\frac{1}{n}\ln(i)}=e^0=1$$
A: Your question seems to be talking about "imaginary" numbers (you actually mean "complex") and "real" numbers as if they are fundamentally different; that all numbers are always one or the other; and that that the is a giant quantuum jump that a "real" number an imaginary number.
Your question seems to be that as $i^{\frac 1n}$ are all one type of number (you said "imaginary" but you seem to mean "non-real complex") and $i^0  = 1$ is the utterly different and incompatible type "real", when what type is $\lim i^{\frac 1n}$?
Well... you have misconceptions.  All numbers are complex and lie on the complex plane.  Numbers that lie exactly on the real axis are "real" numbers and numbers that lie exactly on the imaginary axis are "imaginary" numbers.  (Fun trivia fact!  Zero is the only number that is both real and imaginary!)  All other numbers are neither real nor imaginary bout are just "generic" plain old boring complex numbers.
Now there is nothing earth shaking or shattering about lying on either the imaginary axis or the real axis and the real numbers and the imaginary numbers aren't really any different from their nearby numbers that can be very close to the axis.
Now, limits are about "getting close" so something.  And there's nothing mindblowing or quantuum state altering about a bunch of numbers that "get close" to the real axis but aren't themselves on it.
So $\lim\limits_{n\to \infty}i^{\frac 1n} = 1 \in \mathbb R \subset \mathbb C$ while each $i^{\frac 1n}\not \in \mathbb R$ (but close  to it) is not a weird or surprising result.
==== tl;dr =====
Your question seems to be confusing "imaginary" number with "complex" number and "complex number" with "non-real complex number".
A "complex" number is of the form $a + bi$ where $a$ and $b$ are real.  If $a = 0$ then we get $0 + bi = bi$ and that is called an "imaginary" or "purely" imaginary number.  It is a type of complex number.
If $b = 0$ then $a+0i = a$ and that is .... a real number.  A real number is a type of complex number.  It is not correct to differentiate between complex and real numbers as real numbers are a subset of complex numbers and everything you know (so far) as "numbers" are all complex numbers.
So if $b \ne 0$ so $a + bi$ is not real then we call that a "non real" complex number.
If we define $w^{\frac 1n}$ as the $z$ so that $z^n =w$ there will be $n$ possible values for $z$, Example $i^4, (-i)^4, 1^4, $ and $(-1)^4$ all equal $1$.  So which one is $1^{\frac 14}$?
Or $(\frac {\sqrt 2}2 + \frac {\sqrt 2} 2i)^2 = i$ and $(-\frac {\sqrt 2}2 - \frac {\sqrt 2} 2i)^2$ so which is $i^{\frac 12}$? And for  $z_1= \frac {\sqrt 3}2+ \frac 12  i; z_2 = -\frac {\sqrt{3}}2 + \frac 12 i; z_3 = -i$ then $z_1^3 = z_2^3 = z_3^3 = i$ so which one is $i^{\frac 13}$. 
I'm not actually sure that there is an agreed upon answer to this. But one way of thinking of this is to note all of them lie on a circle equi-angles apart. so we can take $w^{\frac 1n}$ to be first from the real axis in the counter clockwise direction.
Another way of putting the is that for every $z = a + bi \ne 0$ there is an angle $\theta$ and a positive real distance $r$ so that $a + bi = r(\cos \theta + i \sin \theta)$.  (If we are curious $r = \sqrt {a^2 + b^2}$ and $\theta = \arctan \frac ba + k*\pi$ so that $r\sin \theta =b; r\cos \theta = a$)
Using this notation and then $w = a + bi = r*\cos \theta + r*\sin \theta i$ then $w^{\frac 1n} = \sqrt[n]{r}(\cos \frac {\theta}n + i\sin \frac{\theta}n)$.
And so $i^{\frac 1n} = \cos(\frac {90^{\circ}}{n}) + i\sin (\frac {90^{\circ}}n)=\cos (\frac {\pi}{2n}) + i\sin (\frac {\pi}{2n})$
which for $n > 1$ is neither "purely imaginary" nor "real".
But it doesn't take much work to see
$\lim\limits_{n\to\infty}i^{\frac 1n}=\lim\limits_{n\to\infty}\cos (\frac {\pi}{2n}) + i\sin (\frac {\pi}{2n}) = \cos (\lim\limits_{n\to\infty}\frac {\pi}{2n}) + i\sin (\lim\limits_{n\to\infty}\frac {\pi}{2n})=\cos 0 + i\sin 0= 1 + 0i = 1$
which is real.
...... ADDENDUM .....
Hmmm.... I think I did a disservice.
Conventionally $w^{\frac 1n}$ is defined to be "multivalued" where it can be any of the $n$ possible values.
When converting to "polar notation" $(\theta, r)$ because $w = r(\cos (\theta + 2k\pi) + i\sin (\theta + 2k\pi)) $ for all integers $k$ we can define $w^{\frac 1n} = $ the set of then $n$ different $\sqrt[n]{r}(\cos (\frac \theta a + \frac {2k\pi}n) + i \sin (\frac \theta a + \frac {2k\pi}n))$.
It still follows the even if we chose different values for $k$ that 
$\lim\limits_{n\to \infty}\sqrt[n]{r}(\cos (\frac \theta a + \frac {2k\pi}n) + i \sin (\frac \theta a + \frac {2k\pi}n)) =1$.
I don't have time to fix my post today but I may fix it by Monday.
