One-sided limits and continuity

One-sided limits

A one-sided limit looks at one side of a point only. Approaching from the left uses inputs just below $a$; approaching from the right uses inputs just above it. In lesson 2.1.1 the two sides always agreed, but sometimes they do not.

Take $f(x) = \dfrac{|x|}{x}$, which is undefined at 0. The absolute value $|x|$ is $x$ when $x \ge 0$ and $-x$ when $x < 0$ (lesson 0.3.3).

So for every positive $x$, $f(x) = \frac{x}{x} = 1$, and for every negative $x$, $f(x) = \frac{-x}{x} = -1$. The table shows the same:

From the left the outputs stay at $-1$; from the right they stay at 1. Each side has a limit, but the two differ.

For $\frac{|x|}{x}$ at 0, the left limit is $-1$ and the right limit is 1. They differ, so $\lim_{x \to 0} \frac{|x|}{x}$ does not exist.

In the explorer, shrink the gap and watch the two points. However close they come to 0, one output stays at $-1$ and the other at 1. Then switch to the hole from lesson 2.1.1 to see two sides that agree.

A jump in a piecewise function

A piecewise function (lesson 1.1.2) can jump where its pieces meet. Take

Just left of 2, only the first piece is used, so the left limit is $2^2 = 4$. Just right of 2, only the second piece is used, so the right limit is $2 - 1 = 1$. The input 2 itself belongs to the second piece, so $f(2) = 1$.

Blow-ups, side by side

A blow-up can go a different way on each side. For $\frac{1}{x}$ near 0, inputs 0.1, 0.01 and 0.001 give 10, 100 and 1000. Inputs $-0.1$, $-0.01$ and $-0.001$ give $-10$, $-100$ and $-1000$.

So $\lim_{x \to 0^+} \frac{1}{x} = \infty$ and $\lim_{x \to 0^-} \frac{1}{x} = -\infty$: it grows without bound to the right and falls without bound to the left. For $\frac{1}{x^2}$ both sides give $\infty$. Either way no number is approached, so neither function has a limit at 0.

At the edge of a domain

Some functions exist on one side of a point only. $\sqrt{x}$ needs $x \ge 0$ (lesson 1.1.1), so at 0 there is no left side to look at. The right limit is all there is: $\lim_{x \to 0^+} \sqrt{x} = 0$.

$\ln x$ needs $x > 0$, and near 0 it falls without bound. It gives $-2.302585$ at 0.1, $-6.907755$ at 0.001 and $-13.815511$ at $10^{-6}$, so $\lim_{x \to 0^+} \ln x = -\infty$.

Continuity

A function is continuous at $a$ when its graph has no break there: you can draw through $a$ without lifting your pen. In terms of limits, the value at $a$ is exactly where the graph is heading.

A function is continuous on an interval when it is continuous at every point of it. Polynomials, $e^x$, $\sin x$ and $\cos x$ are continuous everywhere. $\sqrt{x}$ is continuous for $x \ge 0$ and $\ln x$ for $x > 0$.

Sums, products and ratios of continuous functions are continuous too, except where a bottom is 0. That is why substitution worked in lesson 2.1.1. For example, $\frac{x + 1}{x - 5}$ is continuous at every input except 5.

Three ways to break

• A hole: the limit exists, but $f(a)$ is missing or different. $\frac{x^2 - 1}{x - 1}$ at 1 has limit 2 and no value. Defining $f(1) = 2$ fills the hole, which is why a hole is also called a removable break.
• A jump: both one-sided limits exist but differ, as with $\frac{|x|}{x}$ at 0, or the piecewise $f$ above at 2. No choice of $f(a)$ can fix it.
• A blow-up: the outputs run off to $\infty$ or $-\infty$ on at least one side, as with $\frac{1}{x}$ at 0.

Joining the pieces

You can choose a piecewise function's numbers so the pieces meet. Take $f(x) = x + k$ for $x < 1$ and $3x$ for $x \ge 1$. The left limit at 1 is $1 + k$, and the right limit and the value are both $3 \times 1 = 3$.

They are equal when $1 + k = 3$, so $k = 2$ makes $f$ continuous. With $k = 2$, the inputs 0.999, 1 and 1.001 give 2.999, 3 and 3.003, with no jump.

Why bisection needs continuity

Bisection (lesson 1.3.2) keeps the half of an interval where the sign of $f$ changes. It relies on one fact: a function that is continuous on $[a, b]$ and changes sign must cross 0 in between. A graph without breaks cannot get from below the axis to above it without crossing.

Without continuity the fact fails. For $\frac{1}{x}$ on $[-1, 2]$, $f(-1) = -1$ and $f(2) = 0.5$ have opposite signs, yet $\frac{1}{x}$ is never 0. Bisection still runs: it closes in on 0, the blow-up, where $\frac{1}{x}$ is huge.

Continuous does not mean smooth. $|x|$ is continuous at 0, since both one-sided limits and the value are 0, but it has a corner there. Lesson 2.2.1 uses one-sided limits to find slopes, and lesson 2.2.2 shows why the corner has none.

Common mistakes

1. Check both sides. A limit from one side says nothing about the other. The limit exists only when the two agree.
2. The value is a separate question. For the piecewise $f$, $f(2) = 1$ matches the right limit, yet the limit at 2 does not exist. Find the piece that holds $a$ itself, as in lesson 1.1.2.
3. $a^-$ means "from below $a$", not "negative". $x \to 2^-$ uses inputs like 1.99, which are positive.
4. A limit can exist where the function is not continuous. A hole has a limit but no value, or the wrong value.
5. $\infty$ from both sides is still no limit. $\frac{1}{x^2}$ grows without bound on both sides of 0, and $\infty$ is not a number.
6. A sign change proves a root only for a continuous function. $\frac{1}{x}$ changes sign across 0 without ever being 0.