$\displaystyle \lim_{x -> 0} (5x - \frac{4}{x})$
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?
$\displaystyle \lim_{x -> 0} (5x - \frac{4}{x})$
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.
Supposing $\displaystyle f$ defined in a punctured neighbourhood of $\displaystyle a$:
$\displaystyle \displaystyle\lim_{x \to a}{f(x)}=+\infty \Leftrightarrow \forall{M>0}\;\exists{ \epsilon >0}:\textrm{\;if\;}0<|x-a|< \epsilon \textrm{\;then\;} f(x)>M$
$\displaystyle \displaystyle\lim_{x \to a}{f(x)}=\infty \Leftrightarrow \forall{M>0}\;\exists{ \epsilon >0}:\textrm{\;if\;}0<|x-a|< \epsilon \textrm{\;then\;}| f(x)|>M$
Our case $\displaystyle |f(x)|=|5x-4/x|\to +\infty$ as $\displaystyle x\to 0$ (two sided) so, satisfies the second definition.
Strictly speaking, that limit does not exist, even as "plus infinity" or "negative infinity"! The two one sided limits are
$\displaystyle \lim_{x\to 0^+} f(x)= -\infty$ (for example, if x= 0.0001, $\displaystyle f(x)= f(0.0001)$$\displaystyle = 5(.0001)- \frac{4}{.0001}$$\displaystyle = .0005- 40000= -39999.9995$ and
$\displaystyle \lim_{x\to 0^-}f(x)= +\infty$ (for example, if x= -0.0001, $\displaystyle f(x)= f(-0.0001)= 5(-.0001)- \frac{4}{-.0001}$[tex]= -.0005+ 40000= 39999.9995.
I was sorely tempted to ask if you understood why $\displaystyle -5\ne 5$!I don't understand why $\displaystyle -\infty\ne\infty$.
Of course, neither "$\displaystyle \infty$" nor "$\displaystyle -\infty$" is actually a number. We are really talking about limits of functions or sequences. $\displaystyle -\infty$ and $\displaystyle \infty$ are different because sequence that converge to each, such as 1, 2, 3, 4, ..., n, ..., which has limit $\displaystyle \infty$, and -1, -2, -3, -4, ..., -n, ..., which has limit $\displaystyle -\infty$, are clearly getting farther apart so cannot converge to the same thing.
(From a "topological" viewpoint, adding $\displaystyle \infty$ and $\displaystyle -\infty$ "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.)
When we compactify $\displaystyle \mathbb{C}$ adding only one point to obtain an homeomorphism with $\displaystyle S^2$ , we denote the added point to $\displaystyle \mathbb{C}$ by $\displaystyle \infty$ .
When we compactify $\displaystyle \mathbb{R}$ adding only one point to obtain an homeomorphism with $\displaystyle S^1$, how do you denote the added point?