Graphing and Evaluating $\lim_{x\to\infty}(\frac{1-x}{x})^x$ This question is a two-part question. So I have two problems.
Problem 1
Before I was evaluating this equation, I graphed the equation to see how it looked before making into a limit. This is what I got:

As you can see from the graph, the domain appears to be $\{X\in\mathbb{R}|0\le x \leq 1\}$ but that is wrong. As you can clearly see that is clearly wrong since there is a value of $x=2$ but it is not plotted on the graph. Same with $x=3,4,5...etc$. 
What I don't understand is that why is $x=1.5$ is undefined and $x=1.4$ is defined. If I do the calculation by hand, I get the following resultant:
$$(\dfrac{1-x}{x})^x\qquad x=1.4$$ 
$$(\dfrac{1-1.4}{1.4})^{1.4}$$
$$(\dfrac{-0.4}{1.4})^{1.4}$$ 
$$Math\;Error\; appears\; on\; my\; calculator$$
Why does desmos state that at $x=1.4$, there is a value
Problem 2
My second is that I need to predict $\lim\limits_{x\to\infty}(\dfrac{1-x}{x})^x$. They I would do it would take out the highest exponent (by looking for the highest exponent in the denominator) from the numerator and denominator, but there is an exponent on the entire fraction, so I am stuck there. I did research on how to do it but I was unable to find an answer that would help for this question.
 A: The expression is ill defined for $x>1$ (except for integer $x$), because then $(1-x)/x < 0$.
In this situation, one usually defines
$$a^b = \exp\big(b\,\text{Log}(a)\big)$$
where $\text{Log}$ is a complex logarithm.
In case you're not familiar with complex logarithms, here's a summary. $\exp$ is periodic in the complex plane: we have $\exp(2\pi i) = 1$, and hence $\exp(z + 2k\pi i) = \exp(z)$ for all $k\in\Bbb Z$.
This means that the complex exponential does not admit a classical inverse, so the complex logarithm is 'multivalued'.
Indeed, we have
$$\exp^{-1}\left(re^{i\theta}\right) = \{\log(r)\,+i(\theta + 2k\pi)\mid k\in\Bbb Z\}.$$
When doing calculations with the complex logarithm, one chooses a branch.
Usually, the branch cut is done along the negative real axis; this is called the principal branch of the complex logarithm.

That your expression is ill defined it easy to see even if we ignore the underlying nuances of complex numbers.
In general, if $a<0$ and $q \in \Bbb Q\setminus\Bbb Z$, then $a^q$ is not well defined in the reals.
Say $q = m/n$ with $m,n\in\Bbb Z$.
Is $a^q = \sqrt[n]{a^m}$, or is it ${\left(\sqrt[n]a\right)}^m$?
Notice that if $n$ is even, then $\sqrt[n]a$ is not well defined, so the order matters (but ideally it shouldn't).
If we choose to define $a^q = \sqrt[n]{a^m}$, when $m$ is odd then $a^m<0$, so for even $n$ the root is still ill-defined.
And even if $n$ were odd, we still run into problems:
if $q=m/n$, then $q=(km)/(kn)$ for any $k\in\Bbb Z$.
In particular, we may choose $k$ even.
The value or 'well-definedness' of $a^q$ must not depend on a particular representation of $q$.

And after all this, there's the problem of defining $a^b$ for $b>0$ irrational.
This is usually done for $a>0$ via continuity: we show that $a^q$ is continuous on the positive rationals, and define
$$a^b = \lim_{\substack{q\,\in\,\Bbb Q\\q\,\to\,b}} a^q.$$
But if $a<0$, $a^q$ is ill defined on the rationals, so how do you define $a^b$?

If you require $x\in\Bbb N$ as $x\to\infty$, then we have
$${\left(\frac{1-x}x\right)}^x = {(-1)}^x\,{\left(1 - \frac1x\right)}^x. \tag{$*$}$$
A classical limit is $\lim_{n\to\infty}{\left(1 + \frac1n\right)}^n = e$.
Indeed, one often defines $e$ this way.
Now, consider
\begin{align}
{\left(1 + \frac1n\right)}^n \cdot {\left(1 - \frac1n\right)}^n
&=
{\left(1 - \frac1{n^2}\right)}^n \leqslant 1.
\end{align}
By Bernoulli's inequality, we have
$$
{\left(1 - \frac1{n^2}\right)}^n \geqslant 1-\frac1n,
$$
so by the squeeze theorem $\lim_{n\to\infty} {\left(1 + \frac1n\right)}^n \cdot {\left(1 - \frac1n\right)}^n = 1$.
It follows from the algebra of limits that $\lim_{n\to\infty} {\left(1 - \frac1n\right)}^n$ exists and
$$\lim_{n\to\infty} {\left(1 - \frac1n\right)}^n
=
\frac{1}{\lim_{n\to\infty} {\left(1 + \frac1n\right)}^n}
= \frac1e.$$
Hence, even for integer $x$, the expression $(*)$ fails to have a limit: for even $x$, $(*)$ approaches $1/e$, while for odd $x$ it approaches $-1/e$.
