# 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?

• The expression is undefined at $x=4$. – paw88789 Feb 13 '15 at 16:46
• is it undefined, or does it go to infinity, because positive infinity is certainly greater than 0! – trumpetlicks Feb 13 '15 at 16:47
• "when applying 4 into the inequality results in a positive infinity" Nope, it does not "result" in this. – Did Feb 13 '15 at 16:49
• Approaching $4$ from the left yields a large negative value, while approaching $4$ from the right yields a large positive value. Even if you allow infinite limits, this one can't be defined because of the ambiguous sign. If it had been $$\frac{x+3}{(x-4)^2}$$ then $+\infty$ would have been more acceptable. – MPW Feb 13 '15 at 16:59
• First let me note that you use a precise language out of its domain of validity. Second, if I try to decipher what you have in mind, I arrive at the idea that you think we should count x=4 as a solution because the ratio would be +oo when x=4 and because +oo would be nonnegative. But one could equally say that the ratio is -oo, no? Conclusion: better play the math game with the rules of the math game (or, transgress its limits but only if one knows exactly what one is doing). – Did Feb 13 '15 at 17:00

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.

• This answers nicely shows why one cannot extend the defintion, and in fact does not get a positive infinity. – quid Feb 13 '15 at 16:49
• Actually, Im not sure I agree with you on this, if you put 1/0 directly into wolfram alpha, it indeed gives back an infinite result! Thus, it actually does move towards infinity – trumpetlicks Feb 13 '15 at 16:54
• @trumpetlicks for me 1/0 into W|A gives "complex infinity" as result. In other words an unsigned infinity. This is perfectly in line with this answer. – quid Feb 13 '15 at 16:58
• @trumpetlicks I checked that. The symbol is not normal infinity. Look at the ~ upon it. It means something very far in any direction. Not necessarily positive. – Arashium Feb 13 '15 at 16:58
• @trumpetlicks compex infinity not only can refer to $+\infty$ or $-\infty$ but also $(+i \infty)$, $(-i \infty)$ or any combination of them such as $(-\infty +i \infty)$ or $(5 -i \infty)$ – Arashium Feb 13 '15 at 17:30

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 $4$.

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.

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 $]-∞,-3]U]4,∞[$