One-Sided Limits, calculus lesson

A parking garage charges by the started hour. At 1 hour and 59 minutes you pay for two hours, and at 2 hours and 1 minute you pay for three. The price just before the two-hour mark and the price just after it are both perfectly clear, but they are different. One-sided limits describe each side on its own, which is exactly what you need at jumps, at corners of piecewise formulas and at the edge of a domain.

Key terms

Left-hand limit

Definition Let $f$ be defined on an open interval $(c, a)$ for some $c < a$ . We write

\lim_{x \to a^-} f(x) = L

and say the left-hand limit of $f$ at $a$ is $L$ if the values $f(x)$ can be made as close to the real number $L$ as we like by taking $x$ sufficiently close to $a$ with $x < a$ .

In plain words Walk toward $a$ from the left only, and ask which number the outputs settle on.

Example Take $f(x) = \lvert x \rvert / x$ . Every $x < 0$ gives $f(x) = -x/x = -1$ . So $\lim_{x \to 0^-} f(x) = -1$ , even though $f(0)$ is undefined.

Right-hand limit

Definition Let $f$ be defined on an open interval $(a, c)$ for some $c > a$ . We write

\lim_{x \to a^+} f(x) = L

and say the right-hand limit of $f$ at $a$ is $L$ if the values $f(x)$ can be made as close to the real number $L$ as we like by taking $x$ sufficiently close to $a$ with $x > a$ .

In plain words The same idea, coming in from the right.

Example For the same $f(x) = \lvert x \rvert / x$ , every $x > 0$ gives $f(x) = x/x = 1$ , so $\lim_{x \to 0^+} f(x) = 1$ . The two sides disagree, and that disagreement is the jump in the graph at $0$ .

Approaching from the side of the domain

Definition If $f$ is defined on an interval $(a, c)$ but on no interval of the form $(b, a)$ , only the right-hand limit at $a$ can be formed. Likewise, if $f$ is defined on $(c, a)$ but on no interval $(a, b)$ , only the left-hand limit at $a$ can be formed. This happens at the endpoints of a domain such as $[a, b]$ .

In plain words You can only approach a point from a side where the function has values. At the edge of the domain there is just one side to come from.

Example The function $f(x) = \sqrt{4 - x^2}$ has domain $[-2, 2]$ . Near $x = 2$ it exists only for $x < 2$ , and as $x \to 2^-$ the quantity $4 - x^2$ shrinks to $0$ , so

\lim_{x \to 2^-} \sqrt{4 - x^2} = 0.

At $x = -2$ only the right side is available, and $\lim_{x \to -2^+} \sqrt{4 - x^2} = 0$ for the same reason. A table for the left side at $2$ shows the trend.

$x$	$1.9$	$1.99$	$1.999$
$\sqrt{4 - x^2}$	$0.6245$	$0.1997$	$0.0632$

One-sided behavior of piecewise functions

Definition If $f(x) = g(x)$ for every $x$ in some interval $(c, a)$ with $c < a$ and $\lim_{x \to a^-} g(x)$ exists, then $\lim_{x \to a^-} f(x) = \lim_{x \to a^-} g(x)$ . The same holds on the right with an interval $(a, c)$ where $c > a$ . The value $f(a)$ plays no part in either one-sided limit.

In plain words To find a one-sided limit of a piecewise function, use the formula that holds just on that side of the point, and ignore the rule for $x = a$ itself.

Example Let

f(x) = \begin{cases} x^2, & x < 1, \\ 3 - x, & x \ge 1. \end{cases}

On the left of $1$ the formula is $x^2$ , so $\lim_{x \to 1^-} f(x) = 1$ . On the right it is $3 - x$ , so $\lim_{x \to 1^+} f(x) = 2$ . Here $f(1) = 2$ happens to match the right side, but that is a fact about the value, not about the limits.

Check yourself

Let $\lfloor x \rfloor$ be the greatest integer less than or equal to $x$ . Find $\lim_{x \to 2^-} \lfloor x \rfloor$ and $\lim_{x \to 2^+} \lfloor x \rfloor$ .

Show the answer

For $1 \le x < 2$ the value is $1$ , so the left-hand limit is $1$ . For $2 \le x < 3$ the value is $2$ , so the right-hand limit is $2$ .

Find both one-sided limits at $x = 2$ of the function

f(x) = \frac{x^2 - 4}{\lvert x - 2 \rvert}.

Show the answer

For $x > 2$ , $\lvert x - 2 \rvert = x - 2$ and $f(x) = x + 2$ , which approaches $4$ . For $x < 2$ , $\lvert x - 2 \rvert = -(x - 2)$ and $f(x) = -(x + 2)$ , which approaches $-4$ .

Let $f(x) = kx + 1$ for $x < 2$ and $f(x) = x^2$ for $x \ge 2$ . For which $k$ are the two one-sided limits at $2$ equal?

Show the answer

The left-hand limit is $2k + 1$ and the right-hand limit is $4$ . They agree when $2k + 1 = 4$ , so $k = \frac{3}{2}$ .

One-Sided Limits

Theory, part 5

Practice, part 6

Key terms

Left-hand limit

Right-hand limit

Approaching from the side of the domain

One-sided behavior of piecewise functions

Check yourself

How this lesson connects

Builds on

Used later in