Finite and infinite limits
Finite limits
We will start by showing a small summary of the properties of finite limits.
Let's suppose that $\displaystyle\lim_{x \to \pm \infty}{f(x)}=a$ and that $\displaystyle\lim_{x \to \pm \infty}{g(x)}=b$, then also:
- $\displaystyle\lim_{x \to \pm \infty}{f(x) \pm g(x)}=\displaystyle\lim_{x \to \pm \infty}{f(x)} \pm \displaystyle\lim_{x \to \pm \infty}{g(x)}=a \pm b$
- $\displaystyle\lim_{x \to \cdot \infty}{f(x) \pm g(x)}=\displaystyle\lim_{x \to \pm \infty}{f(x)} \cdot \displaystyle\lim_{x \to \pm \infty}{g(x)}=a \cdot b$
- If $b \neq 0$, $\displaystyle\lim_{x \to \cdot \infty}{\frac{f(x)}{g(x)}}=\frac{\displaystyle\lim_{x \to \pm \infty}{f(x)}}{\displaystyle\lim_{x \to \pm \infty}{g(x)}}=\frac{a}{b}$
- If $f(x)>$, $\displaystyle\lim_{x \to \pm \infty}{f(x)^{g(x)}}=\displaystyle\lim_{x \to \pm \infty}{f(x)^{\displaystyle\lim_{x \to \pm \infty}{g(x)}}}=a^b$
- If $n$ odd or if $n$ even and $f(x)\geqslant0 \Rightarrow \displaystyle\lim_{x \to \pm \infty}{\sqrt[n]{f(x)}}=\sqrt[n]{\displaystyle\lim_{x \to \pm \infty}{f(x)}}=\sqrt[n]{a}$
- If $\alpha >0$ and $f(x)>0$, $\displaystyle\lim_{x \to \pm \infty}{ \log_{\alpha}f(x)}=\log_{\alpha}\Big(\displaystyle\lim_{x \to \pm \infty}{f(x)}\Big)=\log_{\alpha}a$
If $\displaystyle\lim_{x \to {+} \infty}{f(x)}=3$ and $\displaystyle\lim_{x \to {+} \infty}{g(x)}=-5$ then:
- $\displaystyle\lim_{x \to {+} \infty}{f(x)+g(x)}=3-5=-2$
- $\displaystyle\lim_{x \to {+} \infty}{f(x)-g(x)}=3-(-5)=8$
- $\displaystyle\lim_{x \to {+} \infty}{f(x) \cdot g(x)}=3 \cdot (-5)=-15$
- $\displaystyle\lim_{x \to {+} \infty}{f(x)^g(x)}=3^{-5}=\frac{1}{3^5}=\frac{1}{243}$
- $\displaystyle\lim_{x \to {+} \infty}{g(x)^{f(x)}}$ does not exist since $g(x) < 0$ for $x$ large enough.
- $\displaystyle\lim_{x \to {+} \infty}{\sqrt[3]{g(x)}}=\sqrt[3]{-5}=-\sqrt[3]{5}$
- $\displaystyle\lim_{x \to {+} \infty}{\sqrt{g(x)}}$ does not exist since $g(x) < 0$ for $x$ large enough.
Infinite limits
Let's start by defining what an infinite limit of a function is $f(x)$:
$$\displaystyle\lim_{x \to {+} \infty}{f(x)}=+\infty \Longleftrightarrow \mbox{ given any } k, \mbox{ there exists another number } h$$
$$\mbox{ such that if } x>h \mbox{ then } f(x)>k$$
Intuitively, it means that we can have $f(x)$ as big as we want by choosing a sufficiently large $x$.
Similarly, we define:
$$\displaystyle\lim_{x \to {+} \infty}{f(x)}=-\infty \Longleftrightarrow \mbox{ given any } k, \mbox{ there exists another number } h$$
$$\mbox{ such that if } x>h \mbox{ then } f(x)<-k$$
and for limits when $x$ goes to minus infinity:
$$\displaystyle\lim_{x \to {-} \infty}{f(x)}=+\infty \Longleftrightarrow \mbox{ given any } k, \mbox{ there exists a another number } h$$
$$\mbox{ such that if } x<-h \mbox{ then } f(x)>k$$
$$\displaystyle\lim_{x \to {-} \infty}{f(x)}=-\infty \Longleftrightarrow \mbox{ given any } k, \mbox{ there exists a another number } h $$
$$\mbox{ such that if } x<-h \mbox{ then } f(x)<-k$$
Let's see three basic examples of functions that tend to infinity:
$k$th power: if $k>0, \displaystyle\lim_{x \to {+}\infty}{x^k}=+\infty$
and in particular $\displaystyle\lim_{x \to {+}\infty}{p\cdot x^k}=sign(p) \cdot \infty$ , where $p$ is a real value other zero.
From this point, we deduce that the polynomial functions tend to infinity as $x$ becomes larger.
In this example we can see the function $f(x)=3x^4$. When $x$ becomes large, the function grows to infinity.
Exponential: if $a>1, \displaystyle\lim_{x \to {+}\infty}{a^x}=+\infty$
and likewise if $a>1, \displaystyle\lim_{x \to {+}\infty}{p \cdot a^x}=sign(p) \cdot \infty$ .
An example for this case is the function $f(x)=\dfrac{1}{2}e^x$. It tends to infinity as $x$ tends to infinity.
Logarithmic: if $a>1, \displaystyle\lim_{x \to {+}\infty}{\log_{a}x}=+ \infty$
Similarly if $a>1 \displaystyle\lim_{x \to {+}\infty}{\log_{a}x}=sign(p) \cdot \infty$.
For example, the function $f(x)= \log_{e}x=\ln x$. This function tends to infinity as $x$ becomes very large.
Infinite's arithmetic
Let's suppose that $\displaystyle\lim_{x \to {+} \infty}{f(x)}=+\infty$ and that $\displaystyle\lim_{x \to {+} \infty}{g(x)}=+\infty$, then we observe without problems that:
$$\displaystyle\lim_{x \to {+} \infty}{f(x)+g(x)}=\displaystyle\lim_{x \to {+} \infty}{f(x)}+\displaystyle\lim_{x \to {+} \infty}{g(x)}=+\infty + \infty=+\infty$$
$$\displaystyle\lim_{x \to {+} \infty}{f(x) \cdot g(x)}=\displaystyle\lim_{x \to {+} \infty}{f(x)} \cdot \displaystyle\lim_{x \to {+} \infty}{g(x)}=(+\infty) \cdot (+\infty)=+\infty$$
However, we will have problems when we encounter situations like the following one:
$$\displaystyle\lim_{x \to {+} \infty}{f(x)-g(x)}=\displaystyle\lim_{x \to {+} \infty}{f(x)}-\displaystyle\lim_{x \to {+} \infty}{g(x)}=(+\infty)-(+\infty)$$
since to if we subtract infinity from infinity it gives us an indeterminacy.
Similarly, we might ask ourselves about these properties when we a function with an infinite limit and one with a finite limit.
Let's see a small table that will show us how to work when we have different kinds necessary to produce infinity with other infinites and with finite limits:
| SUMS | PRODUCTS |
| $(+\infty)+a=+\infty$ | $(+\infty)\cdot(+\infty)=+\infty$ |
| $(+\infty)+(+\infty)=+\infty$ | $(+\infty)\cdot(-\infty)=-\infty$ |
| $(-\infty)+a=-\infty$ | $(+\infty)\cdot a=sign(a) \cdot \infty$ |
| $(-\infty)+(-\infty)=-\infty$ | $(-\infty)\cdot a=-sign(a) \cdot \infty$ |
| $-(-\infty)=+\infty$ | |
| DIVISIONS | POWERS |
| $\frac{a}{\pm \infty}=0$ | $(+\infty)^{+\infty}=+\infty$ |
| $\frac{a}{0}=\pm \infty$ if $a\neq 0$ | $(+\infty)^{-\infty}=0$ |
| $\frac{\pm \infty}{0}=\pm \infty$ | if $a$ > $0$ $$(+\infty)^a= + \infty$$ |
| $\frac{0}{\pm \infty}=0$ | if $a$ < $0$ $$(+\infty)^a=0$$ |
| if $a\neq 0$ $$a^0=1$$ | |
| if $a$ > $1$ $$a^{+\infty}=+\infty \\ a^{-\infty}=0$$ | |
| if $0$ <$a$ < $1$ $$a^{+\infty}=0 \\ a^{-\infty}=+ \infty$$ |
These operations can be realized after finding the limits of the functions involved.
Nevertheless, the operations that are not in the table can produce indeterminacies, for example, the following expressions:
$$(+\infty)-(+\infty) \\ 0 \cdot (\pm \infty) \\ \frac{0}{0} \\ (+\infty)^0 \\ 1^{\pm \infty} \\ 0^0 \\ \frac{\pm \infty}{\pm \infty}$$