Calculus · Functions and limits · Lesson 3

One-sided limits and jumps

Walk toward a point from the left, then from the right. If you arrive at the same height both ways, the limit exists. If not, you’ve found a jump.

A function with a step in it

Real rules often switch at a threshold: a tax band, a postage price, a thermostat. Here is a made-up one with a switch at $x = 2$:

$$f(x) = \begin{cases} x + 1 & x < 2 \\ x^2 & x \ge 2 \end{cases}$$

Drag the slider to shrink the distance $d$ from 2. One dot walks in from the left, the other from the right.

from the left, $f(2-d)$
from the right, $f(2+d)$

From the left, the values head to 3. From the right, they head to 4. They never agree.

What height does $f$ approach as $x$ comes toward 2 from the left?

For $x < 2$ the rule is $x + 1$. As $x$ gets close to 2 that heads to $2 + 1 = 3$.

The notation

A small sign on the $a$ says which side you come from:

$$\lim_{x\to 2^-} f(x) = 3 \qquad \lim_{x\to 2^+} f(x) = 4$$

$2^-$ means “from below 2” (the left); $2^+$ means “from above 2” (the right). To compute each, use only the piece of the rule that applies on that side.

Here $f(2) = 2^2 = 4$, the right-hand limit is 4, and the left-hand limit is 3. What is $\displaystyle\lim_{x\to 2} f(x)$?

The two sides disagree (3 vs 4), so there is no single height the function settles toward. And $f(2) = 4$ doesn’t change that: as in lesson 1, the value at the point never decides the limit. It only tells you where the filled dot sits.

Value at the point vs. limit

Three separate questions, three separate answers for our $f$:

It’s tempting to read $f(2)$ and call that the limit. Lesson 1 showed a limit can exist where $f(a)$ doesn’t; here is the reverse: $f(a)$ exists but the limit doesn’t.

A jump made of algebra: $\dfrac{|x|}{x}$

Absolute value is secretly a two-piece function: $|x| = x$ when $x \ge 0$, and $|x| = -x$ when $x < 0$ (making a negative number positive). Tap through:

$\displaystyle\lim_{x\to 0}\frac{|x|}{x}$

  1. Plug in: $\frac{0}{0}$. But nothing factors, so split by side.

  2. Right side, $x > 0$: $|x| = x$, so $\frac{|x|}{x} = \frac{x}{x} = 1$. Every value is 1, so $\lim_{x\to0^+} = 1$.

  3. Left side, $x < 0$: $|x| = -x$, so $\frac{|x|}{x} = \frac{-x}{x} = -1$. So $\lim_{x\to0^-} = -1$.

  4. $1 \ne -1$, so the two-sided limit does not exist. The graph is a flat line at $-1$ that jumps to a flat line at $1$.

$\displaystyle\lim_{x\to 3^-}\frac{|x-3|}{x-3}$

From the left, $x < 3$, so $x - 3$ is negative and $|x-3| = -(x-3)$. Then $\frac{-(x-3)}{x-3} = -1$ for every such $x$, so the left limit is $-1$.

Closing the gap

Often you get to choose a constant so the pieces meet. Slide $k$ until the left piece lands where the right piece starts.

$$g(x) = \begin{cases} kx + 1 & x < 1 \\ x^2 + 3 & x \ge 1 \end{cases}$$

left limit, $k\cdot1 + 1$
right limit, $1^2 + 3$

Which $k$ makes $\displaystyle\lim_{x\to1} g(x)$ exist? (Set the two one-sided limits equal and solve.)

Left limit: $k(1) + 1 = k + 1$. Right limit: $1^2 + 3 = 4$.

Set equal: $k + 1 = 4$. Subtract 1 from both sides: $k = 3$.

One-sided blow-ups

Sides matter for blow-ups too. Near 0, $\frac1x$ is huge and positive just to the right ($\frac{1}{0.001} = 1000$) and huge and negative just to the left ($\frac{1}{-0.001} = -1000$). We write $\lim_{x\to0^+}\frac1x = +\infty$ and $\lim_{x\to0^-}\frac1x = -\infty$. “$= \infty$” is shorthand for “grows without bound”, not a number. Lesson 4 is all about these.

Check

$\lfloor x \rfloor$ (“floor of $x$”) rounds down to a whole number: $\lfloor 2.7 \rfloor = 2$, $\lfloor 3 \rfloor = 3$. What is $\displaystyle\lim_{x\to 3^-}\lfloor x \rfloor$?

Just left of 3, $x$ is something like 2.99, and $\lfloor 2.99 \rfloor = 2$. Every value on that side is 2, so the left limit is 2, even though $\lfloor 3 \rfloor = 3$. (The right limit is 3, so the two-sided limit at 3 doesn’t exist: a staircase has a jump at every whole number.)

$h(x) = \dfrac{x^2-4}{x-2}$ for $x < 2$, and $h(x) = 3x - 2$ for $x > 2$. Find $\displaystyle\lim_{x\to2} h(x)$.

Left: $\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x + 2 \to 4$ (lesson 2’s factoring move).

Right: $3(2) - 2 = 6 - 2 = 4$.

Both sides give 4, so the limit is 4, even though $h(2)$ isn’t defined at all.

$\displaystyle\lim_{x\to 0^-}\frac{1}{x}$ is…

Left of 0, $x$ is a tiny negative number, and 1 divided by a tiny negative is a huge negative: $\frac{1}{-0.01} = -100$. So it falls without bound.

In one or two sentences: why can’t $\lim_{x\to a} f(x)$ exist if the left and right limits are different?

The limit has to be a single number $L$ that $f(x)$ gets close to for every $x$ near $a$, on both sides. If the left side crowds toward 3 and the right toward 4, no single $L$ is close to both, so there’s no limit.