# Positive or Negative Infinity?

• Feb 26th 2011, 09:32 AM
RogueDemon
Positive or Negative Infinity?
$\lim_{x -> 0} (5x - \frac{4}{x})$

Is the answer to this question positive or negative infinity? The calculator at Online Limit Calculator gives positive infinity, but wouldn't -4/0 be negative?
• Feb 26th 2011, 09:44 AM
Plato
Quote:

Originally Posted by RogueDemon
$\lim_{x -> 0} (5x - \frac{4}{x})$ Is the answer to this question positive or negative infinity? The calculator at Online Limit Calculator gives positive infinity, but wouldn't -4/0 be negative?

Have a look at this.
It appears that it depends on the side of 0 we choose.
• Feb 26th 2011, 10:03 AM
FernandoRevilla
Quote:

Originally Posted by RogueDemon
$\lim_{x -> 0} (5x - \frac{4}{x})$

Is the answer to this question positive or negative infinity? The calculator at Online Limit Calculator gives positive infinity, but wouldn't -4/0 be negative?

The calculator is correct. Take into account that $\infty\neq +\infty$ . Look at the three posibilities: two sided, plus and minus.
• Feb 26th 2011, 10:35 AM
NOX Andrew
I don't understand why $\infty \neq +\infty$.
• Feb 26th 2011, 10:49 AM
FernandoRevilla
Quote:

Originally Posted by NOX Andrew
I don't understand why $\infty \neq +\infty$.

Supposing $f$ defined in a punctured neighbourhood of $a$:

$\displaystyle\lim_{x \to a}{f(x)}=+\infty \Leftrightarrow \forall{M>0}\;\exists{ \epsilon >0}:\textrm{\;if\;}0<|x-a|< \epsilon \textrm{\;then\;} f(x)>M$

$\displaystyle\lim_{x \to a}{f(x)}=\infty \Leftrightarrow \forall{M>0}\;\exists{ \epsilon >0}:\textrm{\;if\;}0<|x-a|< \epsilon \textrm{\;then\;}| f(x)|>M$

Our case $|f(x)|=|5x-4/x|\to +\infty$ as $x\to 0$ (two sided) so, satisfies the second definition.
• Feb 27th 2011, 05:55 AM
HallsofIvy
Strictly speaking, that limit does not exist, even as "plus infinity" or "negative infinity"! The two one sided limits are
$\lim_{x\to 0^+} f(x)= -\infty$ (for example, if x= 0.0001, $f(x)= f(0.0001)$ $= 5(.0001)- \frac{4}{.0001}$ $= .0005- 40000= -39999.9995$ and
$\lim_{x\to 0^-}f(x)= +\infty$ (for example, if x= -0.0001, $f(x)= f(-0.0001)= 5(-.0001)- \frac{4}{-.0001}$[tex]= -.0005+ 40000= 39999.9995.

Quote:

I don't understand why $-\infty\ne\infty$.
I was sorely tempted to ask if you understood why $-5\ne 5$!(Rofl)

Of course, neither " $\infty$" nor " $-\infty$" is actually a number. We are really talking about limits of functions or sequences. $-\infty$ and $\infty$ are different because sequence that converge to each, such as 1, 2, 3, 4, ..., n, ..., which has limit $\infty$, and -1, -2, -3, -4, ..., -n, ..., which has limit $-\infty$, are clearly getting farther apart so cannot converge to the same thing.

(From a "topological" viewpoint, adding $\infty$ and $-\infty$ "compactifies" the real number line making it topologically equivalent to a closed interval. Given any topological space, we can always find its "one point compactification", adding a single point, and "open" sets containing that one point, to make it compact- However, it changes the topological properties of the space much more than the "Stone-Chech compactification" above. If we did that with the set of real numbers, we would get a single "point at infinity" but now the it would be topologically equivalent to a circle rather than a line segment.)
• Feb 27th 2011, 06:34 AM
NOX Andrew
You seem to have misquoted me. I understand the difference between negative infinity and infinity, just not the difference between positive infinity and infinity.
• Feb 27th 2011, 06:44 AM
FernandoRevilla
Quote:

Originally Posted by HallsofIvy
Strictly speaking, that limit does not exist, even as "plus infinity" or "negative infinity"!

When we compactify $\mathbb{C}$ adding only one point to obtain an homeomorphism with $S^2$ , we denote the added point to $\mathbb{C}$ by $\infty$ .

When we compactify $\mathbb{R}$ adding only one point to obtain an homeomorphism with $S^1$, how do you denote the added point?