Math Core

Estimating limits from tables and graphs builds intuition, but it is slow and never certain. The limit laws let you compute limits exactly by breaking a complicated function into simple pieces whose limits you already know, then reassembling the answer.

Two basic limits

Every limit law computation eventually rests on two facts that are clear from the graphs of $y = k$ and $y = x$ :

\lim_{x \to c} k = k \qquad \text{and} \qquad \lim_{x \to c} x = c.

The first says a constant function stays at height $k$ no matter where $x$ goes. The second says that as $x$ approaches $c$ , the output of the identity function, which is just $x$ , also approaches $c$ .

The limit laws

Limit laws

Suppose $\displaystyle \lim_{x \to c} f(x) = L$ and $\displaystyle \lim_{x \to c} g(x) = M$ , where $L$ and $M$ are real numbers, and let $k$ be a constant. Then:

Law	Statement
Sum	$\displaystyle \lim_{x \to c} [f(x) + g(x)] = L + M$
Difference	$\displaystyle \lim_{x \to c} [f(x) - g(x)] = L - M$
Constant multiple	$\displaystyle \lim_{x \to c} [k \, f(x)] = kL$
Product	$\displaystyle \lim_{x \to c} [f(x) \, g(x)] = LM$
Quotient	$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M}$ , provided $M \ne 0$
Power	$\displaystyle \lim_{x \to c} [f(x)]^n = L^n$ for any positive integer $n$
Root	$\displaystyle \lim_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{L}$ (for even $n$ , require $L > 0$ )

In words: the limit of a sum is the sum of the limits, the limit of a product is the product of the limits, and so on, as long as the individual limits exist (and, for quotients, the denominator's limit is not zero). The same laws hold for one-sided limits.

The laws make sense intuitively. If $f(x)$ is close to $L$ and $g(x)$ is close to $M$ , then $f(x) + g(x)$ is close to $L + M$ and $f(x) g(x)$ is close to $LM$ . A formal proof uses the precise definition of a limit, which is beyond the AP course.

Worked example: Using given limits

Suppose $\displaystyle \lim_{x \to 2} f(x) = 5$ and $\displaystyle \lim_{x \to 2} g(x) = -3$ . Evaluate

\lim_{x \to 2} \frac{3f(x) - g(x)^2}{f(x) + g(x)}.

Solution. First check the denominator: $\displaystyle \lim_{x \to 2} [f(x) + g(x)] = 5 + (-3) = 2 \ne 0$ , so the quotient law applies. The numerator's limit is

3 \cdot 5 - (-3)^2 = 15 - 9 = 6.

Therefore the limit is $\dfrac{6}{2} = 3$ .

Direct substitution

Applying the laws over and over to a polynomial gives a powerful shortcut. For example,

\lim_{x \to 3} (2x^2 - 5x + 1) = 2 \left(\lim_{x \to 3} x\right)^2 - 5 \lim_{x \to 3} x + \lim_{x \to 3} 1 = 2(9) - 15 + 1 = 4,

which is exactly what you get by plugging in $x = 3$ .

Direct substitution property

If $p$ is a polynomial, then $\displaystyle \lim_{x \to c} p(x) = p(c)$ for every real $c$ .

If $r(x) = \dfrac{p(x)}{q(x)}$ is a rational function and $q(c) \ne 0$ , then $\displaystyle \lim_{x \to c} r(x) = r(c)$ .

The same is true for the other functions you know well, at every point in their domains: roots, $\sin x$ , $\cos x$ , $\tan x$ , $e^x$ , $\ln x$ , and so on. Functions with this property are called continuous, and you will study them carefully later in this unit. For now, the practical rule is: if substituting $c$ gives a real number, that number is the limit.

Worked example: Direct substitution with a trigonometric function

Evaluate $\displaystyle \lim_{x \to \pi} \frac{x \cos x}{2 + \sin x}$ .

Solution. The denominator at $x = \pi$ is $2 + \sin \pi = 2 \ne 0$ , so substitute:

\frac{\pi \cos \pi}{2 + \sin \pi} = \frac{\pi(-1)}{2 + 0} = -\frac{\pi}{2}.

Composite functions

A limit can pass inside an outer function when that outer function is continuous at the inner limit.

Limit of a composition

If $\displaystyle \lim_{x \to c} g(x) = L$ and $f$ is continuous at $L$ , then

\lim_{x \to c} f(g(x)) = f\left(\lim_{x \to c} g(x)\right) = f(L).

For example, $\displaystyle \lim_{x \to 1} \sqrt{x^2 + 8} = \sqrt{\lim_{x \to 1} (x^2 + 8)} = \sqrt{9} = 3$ .

Common mistake

The limit laws only apply when the individual limits exist. You cannot split $\displaystyle \lim_{x \to 0} \left[\frac{1}{x} - \frac{1}{x}\right]$ into $\displaystyle \lim_{x \to 0} \frac{1}{x} - \lim_{x \to 0} \frac{1}{x}$ , because neither piece has a limit. (The original limit is $0$ , since the expression equals $0$ for all $x \ne 0$ .) In the other direction, a sum can have a limit even when its pieces do not, so "the pieces have no limit" never proves "the sum has no limit."

Worked example: Limits from graphs, combined

The graphs of $f$ and $g$ are shown. Find $\displaystyle \lim_{x \to 1} [f(x) + g(x)]$ .

The graph of f: open circles at (1, 1) and (1, −1).Open in grapher →

The graph of g: open circles at (1, 1) and (1, 3).Open in grapher →

Solution. Neither $f$ nor $g$ has a two-sided limit at $1$ , so you must work with one-sided limits.

From the left: $f \to 1$ and $g \to 1$ , so $f + g \to 2$ .

From the right: $f \to -1$ and $g \to 3$ , so $f + g \to 2$ .

Both one-sided limits of the sum equal $2$ , so $\displaystyle \lim_{x \to 1} [f(x) + g(x)] = 2$ , even though neither individual limit exists.

Tip

When the quotient law fails because the denominator's limit is $0$ , the limit is not automatically undefined. It is a signal to do more work: simplify algebraically (next lesson) or check for a vertical asymptote.

Practice

Practice 1

Suppose $\displaystyle \lim_{x \to 4} f(x) = 2$ and $\displaystyle \lim_{x \to 4} g(x) = -3$ . Find $\displaystyle \lim_{x \to 4} [f(x) \, g(x)]$ .