07 Other Kinds of Limit
Section 2.4 covers one-sided limits, limits at infinity, and infinite limits. All of these types of limit are less important for us than the basic limit of a function at a point from the previous reading. Nevertheless, you should know the definitions and be able to work with them. We will be most interested in $\ds\lim_{x\to+\infty}f(x)$ because of its analogy with limits of sequences.
The limit from the left of a function $f$ at a point $a$ is denoted $\ds\lim_{x\to a^-}f(x)$ and is defined as follows: $\ds\lim_{x\to a^-}f(x)=L$ if for every $\eps>0$, there is a $\delta>0$ such that for every $x$ satisfying $-\delta<x-a<0$, $f(x)$ is defined and $|f(x)-L|<\eps.$ Note that for this limit to be defined, $f$ must be defined on some open interval, $(c,a)$, to the left of $a.$ The limit from the right is defined similarly: $\ds\lim_{x\to a^+}f(x)=L$ if for every $\eps>0$, there is a $\delta>0$ such that for every $x$ satisfying $0<x-a<\delta$, $f(x)$ is defined and $|f(x)-L|<\eps.$
Probably the most important thing about one-sided limits is the fact that $\ds\lim_{x\to a}f(x)=L$ exists if and only if both one sided limits at $a$ exist and are equal to $L.$ This is clear from the definitions and the fact that $|x-a|<\delta$ if and only if $-\delta<x-a<\delta.$
$\ds\lim_{x\to a}f(x)=L$ means, informally, that when $x$ is "near" $a$, then $f(x)$ is "near" $L.$ The epsilon-delta definition makes this idea rigorous. When thinking about infinite limits and limits at infinity, the question is, What does it mean for a number to be "near" infinity? Being near $+\infty$ should mean being greater than some number $M$ (where $M$ is thought of as being big in the same way that $\eps$ is thought of as being small). Being near $-\infty$ should mean being less than $-M$ for some big number $M.$
So, $\ds\lim_{x\to+\infty}f(x)=L$, which means informally that $f(x)$ is near $L$ when $x$ is near $+\infty,$ is defined rigorously as follows: $\ds\lim_{x\to+\infty}f(x)=L$ if for every $\eps>0$, there is an $M$ such that for every $x> M,$ $f(x)$ is defined and $|f(x)-L|<\eps.$ Similarly, $\ds\lim_{x\to a}f(x)=+\infty$ is defined as follows: $\ds\lim_{x\to a}f(x)=+\infty$ if for every $M$, there is a $\delta>0$ such that for every $x$ satisfying $0<|x-a|<\delta$, $f(x)>M.$ Limits involving $-\infty$ are handled similarly.
It is important to note that writing $\ds\lim_{x\to a}f(x)=+\infty$ is an example of abuse of notation, since $\ds\lim_{x\to a}f(x)=+\infty$ means that the limit does not exist. but the notation makes it looks like the limit does exist. Of course, infinity is not a number so saying that the limit is infinite really means that the limit does not exist because the function grows larger than any finite bound as $x$ gets closer to $a.$
(I will sometimes write $\infty$ when I should be writing $+\infty,$ but $+\infty$ is more correct to avoid any ambiguity.)
One sided limits and limits at infinity have pretty much all of the same properties as basic limits, and the proofs are easy adaptations of the proofs in the basic case. I will do one of the more complex proofs as an example.
Theorem: If $\ds\lim_{x\to+\infty}f(x)=L$ and $\ds\lim_{x\to+\infty}g(x)=K$ and $K\ne 0$, then ${\ds\lim_{x\to+\infty}}\frac{f(x)}{g(x)}=\frac LK.$
Proof: Let $\eps>0$. We want to find $M$ such that $x> M$ implies $\frac{f(x)}{g(x)}$ is defined and $\big|\frac{f(x)}{g(x)}-\frac LK\big|<\eps.$
Since the limits at $+\infty$ of $f$ and $g$ exist and $K\ne0,$ We can find an $M_1$ such that for $x> M_1,$ $f(x)$ and $g(x)$ both exist, and $|g(x)|>\frac{|K|}{2}.$ [For the last inequality, since $\ds\lim_{x\to+\infty}g(x)=K$ and $\frac{|K|}{2}>0$, there is an $N$ such that for $x> N$, $|g(x)-K|<\frac{|K|}{2}.$ This implies that for $x> N,$ $|g(x)|>\frac{|K|}{2}$. See the next-to-last theorem in the fifth reading guide.]
Note that when $x>M_1$, we have $|g(x)|>\frac{|K|}2,$ which in turn implies $\frac{1}{|g(x)|}<\frac2{|K|}$. Using $\ds\lim_{x\to+\infty}f(x)=L,$ choose $M_2$ such that for any $x>M_2$, $|f(x)-L|<|K|\frac\eps4.$ Using $\ds\lim_{x\to+\infty}g(x)=K,$ choose $M_3$ such that for any $x>M_3$, $|g(x)-K|<\frac{|K|^2\eps}{4(|L|+1)}.$ [Obviously, it took a lot of looking ahead to figure out what values were needed here!] Let $M=\max(M_1,M_2,M_3).$ Then for any $x>M,$ $\frac{f(x)}{g(x)}$ is defined, and $$\begin{align*} \left|\frac{f(x)}{g(x)} - \frac LK\right| & = \left|\frac{Kf(x)-Lg(x)}{Kg(x)}\right| \\[8pt] & = \frac{1}{|Kg(x)|}\cdot |Kf(x)-Lg(x)| \\[8pt] & = \frac{1}{|Kg(x)|}\cdot |(Kf(x)-KL)+(KL-Lg(x))| \\[8pt] & \le \frac{1}{|Kg(x)|}\cdot (|Kf(x)-KL| + |KL-Lg(x)|) \\[8pt] & = \frac{1}{|K|}\cdot \frac{1}{|g(x)|}\cdot (|K|\cdot|f(x)-L| + |L|\cdot|g(x)-K|) \\[8pt] & < \frac{1}{|K|} \cdot \frac{2}{|K|}\cdot \left(|K|\cdot|K|\frac\eps4 + |L|\cdot \frac{|K|^2\eps}{4(|L|+1|)}\right) \\[8pt] & = \frac{2}{|K|^2}\cdot\left(|K|^2\frac\eps4 +\frac{|L|}{|L|+1}|K|^2\frac\eps4\right) \\[8pt] & < 2\cdot (\frac\eps4 + 1\cdot\frac\eps4)\\[8pt] & = \eps \end{align*}$$ which is what we needed to show. $\qed$