Why does $(x+3)/(x-4) \geq 0$ not include 4 in the interval result? Wolfram alpha, the book I am going through, an other sources all give the resulting interval for:
$$
\frac{x+3}{x-4} \geq 0
$$
as:
$$
    (-\infty, -3] \cup (4, \infty).
$$
I am struggling to understand why 4 is not included in the set, when applying 4 into the inequality results in a positive infinity, which is definitively greater than 0?
Can anyone explain this?
 A: How do you know 
$$lim_{x \to 0}\frac1x$$
is $+\infty$ ?
if $x \to 0^-$
then it is $-\infty$
Beside that, $\frac10$ is undefined not $\infty$. Do not confuse it with limit. Any number more than 4 (even with tiny small value higher) matches your example which is fair enough. 
A: The simple answer is that $4$ is not in the domain of the function on the left or you inequality. You can't evaluate that expression at $4$ because you would be dividing by $0$.
Remember that when you are asked to solve
$$
\frac{x+3}{x-4} \geq 0
$$
you are asked for the set of all real numbers $x$ such that when you compute $(3+x)/(x-4)$ then you get a non-negative number. And you simply can't evaluate (or apply as you say) the expression at $4$.
A: The implied domain for:
$$\dfrac {x + 3}{x - 4} \ge 0$$
is a subset of all real numbers $x$ that satisfy the inequality. Whatever convention you define for $\dfrac 7 0$, if you feel you need to create some sort of definition, will yield results inconsistent with the properties of real numbers.
The fact that some computer algebra system returns unsigned infinity for $\dfrac 7 0$ is merely a convention used in some complex analysis settings. Even in those settings, it doesn't behave like a number. With complex numbers, your inequality doesn't make sense anyway, because the complex numbers can't be ordered by $\ge$ in any useful way.
The other answers already address why you can't choose $+\infty$ or $-\infty$, whatever meaning you attach to those two symbols.
A: By Inequality:
Here $4$ is free boundry number and by associated equation solution gives $x=-3$.
So we have three region (-∞,-3),(-3,4),(4,∞).
Region (-∞,-3) and (4,∞) are true. And Region (-3,4) is false. So solution set is $]-∞,-3]U]4,∞[$
