Euler Limit with -1 to infinity? Evaluate $$\lim_{x\to0^{+}}\left(\frac{4}{x-4}\right)^{1/x}$$
If the problem was $$4-x$$ it would be fine, but it is $$x-4$$ so I end up with $$(-1)^{+\infty}$$ I thought it would be divergent because we can approach infinity by odd numbers and even numbers and on that case, the limit could vanish to a positive number or to a negative number. But my surprise was that when I put that on Wolfram it gave me a limit.  
 A: There is almost surely a mistake in the printing.
I assume you are studying the calculus of real numbers.
The function you are taking a limit at is not defined for $x < 4$, since it is in the form of a negative number raised to a power. Even if we decided to use the common extension to allow rational exponents with odd denominator, that still leaves undefined all the places where $x$ is irrational!
While it is possible for some conventions to yet make sense of such a situation, stressing the fine details of such conventions in this sort of way is not something introductory textbooks tend to do.

If you did this calculation in wolfram alpha, then it didn't give you a limit. What it did was interpret the exponentiation as complex exponentiation, 
and then returned to you a whole set of limit points — the complex numbers of the form $z=\exp\left( \frac{1}{4} + 2 \pi \mathbf{i} t \right)$ where $t$ is a real number ranging over the interval $[0, \pi]$. That is, the set of complex numbers with norm $e^{1/4}$ and nonnegative imaginary part.
The limit only exists in the special case that there is a single limit point.
A: Inverting we have,
$$\lim_{x \to 0^+} \left(\frac{x-4}{4}\right)^{1/x}$$
$$=\lim_{x \to 0^+} \left(\frac{x}{4}-1 \right)^{1/x}$$
Interestingly enough if $x \in \mathbb{Q}$ is written as a simplified fraction with a even denominator then we can rewrite the limit in such case as,
$$=\lim_{x \to 0^+} \left(1-\frac{x}{4}\right)^{1/x}$$
$$=e^{-\frac{1}{4}}$$
And so our original limit tends to $e^{1/4}$.
However if $x \in \{ \frac{1}{n} \ \text{where} \ n \in \mathbb{N} \ \text{and is odd} \}$  we may rewrite the limit as,
$$=-\lim_{x \to 0^+} \left(1-\frac{x}{4}\right)^{1/x}$$
And then our original limit will tend to $-e^{1/4}$.
We see the limit doesn't exist.
