The calculation of limits

The idea of a limit asks a very specific question: as the input $x$ of a function is pushed closer and closer to some fixed value $x_0$ (without ever actually landing on $x_0$), what value does the output $f(x)$ get closer and closer to?

Illustration — a polynomial. Take $f(x)=x^2+3$ and study its behaviour near $x=2$. Building a table of values with $x$ approaching $2$ from below ($1.7,1.9,1.95,1.99,\dots$) and from above ($2.3,2.1,2.05,2.01,\dots$) shows $f(x)$ closing in on $7$ from both directions — which is also exactly $f(2)=2^2+3$ (Fig. 9.1 plots the parabola $y=x^2+3$ and marks this behaviour near $x=2$). We say the left limit of $f$ at $2$ equals $7$, written $\lim\limits_{x\to2^-} f(x)=7$, and the right limit also equals $7$, written $\lim\limits_{x\to2^+} f(x)=7$. When the left and right limits agree, the common value is called simply the limit, $\lim\limits_{x\to2} f(x)=7$. The important point of this first illustration is that here the limit could be found by plain substitution — but that is a lucky coincidence of $f$ being a polynomial, not something to rely on in general.

Illustration — a function undefined at the target point. Take $f(x)=\dfrac{x^2-16}{x+4}$, whose domain excludes $x=-4$. Even though $f(-4)$ itself does not exist, the limit as $x\to-4$ can still be investigated, because the phrase "$x\to-4$" only ever considers $x$-values near $-4$, never $x=-4$ itself. Algebraically, for any $x\ne-4$, $f(x)=\dfrac{(x+4)(x-4)}{x+4}=4-x$, so the graph of $f$ is exactly the straight line $y=4-x$ except for a single missing point ("hole" or "puncture") directly above $x=-4$ (Fig. 9.2). As $x$ approaches $-4$ from either side, $f(x)$ approaches $8$, so $\lim\limits_{x\to-4} f(x)=8$ even though $f(-4)$ is undefined. This is the key lesson: whether or not $f$ is even defined at $x_0$ has no bearing whatsoever on whether the limit exists at $x_0$.

Illustration — a limit that fails to exist. Take $f(x)=\dfrac{|x|}{x}$ (undefined at $x=0$). For $x>0$, $f(x)=1$; for $x<0$, $f(x)=-1$ (Fig. 9.3 shows two disconnected open half-lines at heights $\pm1$, meeting nowhere above $x=0$). No matter how close $x$ gets to $0$, there are always nearby points giving $f(x)=1$ and other nearby points giving $f(x)=-1$ — the function never settles on a single value as $x\to0$. Hence $\lim\limits_{x\to0^-} f(x)=-1$ while $\lim\limits_{x\to0^+} f(x)=+1$: the one-sided limits disagree, so $\lim\limits_{x\to0} f(x)$ does not exist. (At any other point, e.g. $x=2$ or $x=-3$, the one-sided limits obviously agree and the ordinary limit does exist.)

These three illustrations converge on a formal description of what a limit is:

Definition 9.1. Let $I$ be an open interval containing $x_0$, and $f:I\to\mathbb R$. We say the limit of $f(x)$ as $x$ approaches $x_0$ is $L$ — written $\lim\limits_{x\to x_0} f(x)=L$ — if, whenever $x$ is taken sufficiently close to $x_0$ from either side (with $x\ne x_0$), $f(x)$ becomes correspondingly close to $L$.

The definition deliberately excludes $x=x_0$ itself, which is exactly why the hole at $x=-4$ in the second illustration does not prevent the limit from existing there.

Worked techniques from Examples 9.1–9.6. A recurring technique in this section is splitting a function built from $|x|$, $\sqrt{x}$, $\log x$, or the greatest-integer function $\lfloor x\rfloor$ into its separate pieces and evaluating one-sided limits on each piece:

• $\lim_{x\to0}|x|=0$: since $|x|=x\to0$ from the right and $|x|=-x\to0$ from the left, both one-sided limits agree.
• $\lim_{x\to0}\sqrt x$: $\sqrt x$ is not even defined for $x<0$, so the left-hand limit cannot be formed at all — hence the two-sided limit does not exist, even though the right-hand limit $\lim_{x\to0^+}\sqrt x=0$ does. Similarly $\log x$ has no left-hand limit as $x\to0^-$, since $\log x$ is undefined for $x\le0$ — a reminder that a one-sided limit needs the function to actually be defined on that side. …