Intuition behind $\lim\limits_{n\rightarrow \infty}x^{1/n} = 1$ A limit identity that blows my mind is the following:
$\lim\limits_{n\rightarrow \infty}x^{1/n} = 1$ 
for   $( x > 0) $ 
Rather than just accepting this limit as a fact, I am wondering whether there is a nice way to prove it (preferably visually).
A huge chunk of my confusion comes from the fact that raising a number to a non whole power ex: $3^{0.35}$ can only be computed through computers.
In a similar manner square rooting a number that is not a perfect square, perfect cube, etc... can only be done through computers! Example: 
Let x = 3 and N = 5 thus, $3^{1/5}$.
Assuming that I cannot fully visualize this, how can I go on to say something like $0.35^{1/250} \approx 1$ 
Edit: Thanks to everyone who helped solve this. With the exception of the guy who claimed "anything raised to zero = one" , I loved all the proofs and comments.
 A: First observe that for any $a > 0$,  $x^{1 \over n} < a$ is equivalent to $x < a^n$. 
Suppose $a > 1$. Then if $n$ is large enough, $x < a^n$ will be satisfied since $a > 1$. So $x^{1 \over n} < a$.
Similarly, if $b < 1$, if $n$ is large enough, $x > b^n$ will be satisfied since $b < 1$. So $x^{1 \over n} > b$.
Letting $b = 1 - \epsilon$ and $a = 1 + \epsilon$ for small $\epsilon > 0$, we see that for $n$ large enough we have 
$$1 - \epsilon <  x^{1 \over n} < 1 + \epsilon$$
Equivalently, for $n$ large enough we have
$$|x^{1 \over n} - 1| < \epsilon$$
This is exactly the statement that $\lim_{n \rightarrow \infty} x^{1 \over n} = 1$.
A: We examine three cases; $(1)$ $x=1$, $(2)$ $x>1$, $(3)$ $x<1$.

Case $\displaystyle (1)$:  $\displaystyle x=1$.
When $x=1$, $x^{1/n}=1$ for all $n\in \mathbb{N}$ and the result is trivial.

Case $\displaystyle (2)$:  $\displaystyle x>1$.
If $x>1$, then for any $n\in \mathbb{N}$, $x^{1/n}>1$.  We define the sequence $y_n$ by the expression 
$$y_n\equiv x^{1/n}-1\tag 1$$ 
where we see that $y_n>0$.  Rearranging $(1)$, we write $x=(1+y_n)^n$ whence from Bernoulli's Inequality we find that 
$$x>1+ny_n\tag2$$
Solving $(2)$ for $y_n$ reveals
$$0<y_n<\frac{x-1}{n}\tag3$$
Application of the squeeze theorem to $(3)$ yields
$$\lim_{n\to\infty}y_n=0$$
from which we find from $(1)$
$$\lim_{n\to\infty}x^{1/n}=1$$ 

Case $\displaystyle (3)$:  $\displaystyle x<1$.
If $x<1$, then for any $n\in \mathbb{N}$, $x^{1/n}<1$.  We define the sequence $z_n$ by the expression 
$$z_n\equiv \frac1{x^{1/n}}-1\tag 4$$ 
where we see that $z_n>0$.  Rearranging $(4)$, we write $x=\frac1{(1+z_n)^n}$ whence from Bernoulli's Inequality we find that 
$$x<\frac1{1+nz_n}\tag5$$
Solving $(5)$ for $z_n$ reveals
$$0<z_n<\frac{1/x-1}{n}\tag6$$
Application of the squeeze theorem to $(6)$ yields
$$\lim_{n\to\infty}z_n=0$$
from which we find from $(4)$
$$\lim_{n\to\infty}x^{1/n}=1$$ 

And we are done!
A: Suppose that the limit of the function is some value $y$. Then, $\lim_{n\rightarrow\infty} x^{1/n} = y$. Take the natural log of both sides. Since the natural log is continuous for positive values, you can pass it through the limit and get 
$$ \lim_{n\rightarrow\infty} \ln{x^{1/n}} = \lim_{n\rightarrow\infty} \frac{\ln{x}}{n} = \ln y.$$
Then, as $n \rightarrow \infty$ you get that $$0 = \ln y.$$ Then, $$e^0 = e^{\ln y} \rightarrow 1 = y.$$
Therefore, $\lim_{n\rightarrow\infty} \ln x^{1/n} = 1$. 
A: Computers
It is not true that "raising a number to a non whole power," like $3^{0.35}$, "can only be computed through computers". Anything that a computer can do can be done with manual computation. One way is by using logarithms and exponentials, tables of which were computed long before computers. Computers just make the computations faster and generally less error prone.

The Limit
Bernoulli's Inequality says that for $n\ge1$ and $x\gt-1$,
$$
1+x\le\left(1+\frac xn\right)^n\tag1
$$
raising to the $1/n$ power
$$
(1+x)^{1/n}\le1+\frac xn\tag2
$$
Furthermore, for $n\ge1$ and $x\gt-1$, so that $\frac{x}{1+x}\lt1$, Bernoulli's Inequality says that
$$
1-\frac{x}{1+x}\le\left(1-\frac{x}{n(1+x)}\right)^n\tag3
$$
raising to the $-1/n$ power
$$
\begin{align}
(1+x)^{1/n}
&\ge1+\frac{x}{n(1+x)-x}\\
&\ge1+\frac{x}{n(1+x)}\tag4
\end{align}
$$
Thus, for $n\ge1$ and $x\gt-1$,
$$
1+\frac{x}{n(1+x)}\le(1+x)^{1/n}\le1+\frac xn\tag5
$$
The Squeeze Theorem applied to $(5)$ says that for $x\gt-1$,
$$
\lim_{n\to\infty}(1+x)^{1/n}=1\tag6
$$
then substituting $x\mapsto x-1$ says that for $x\gt0$,
$$
\lim_{n\to\infty}x^{1/n}=1\tag7
$$
