Limit Laws — AP® Calculus AB Notes

So far you have estimated limits by looking: at a graph, at a table. Now you will find them exactly, with rules. The good news is that the rules are the ones you would guess. Limits respect ordinary arithmetic. If you know where $f(x)$ and $g(x)$ are heading, you know where $f(x) + g(x)$ , $f(x)g(x)$ and many other combinations are heading. For most functions in this course, the rules boil down to one step: substitute.

The limit laws

Suppose $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ , and both of these limits exist. Then:

Law	In symbols	Condition
Sum and difference	$\lim_{x \to a} \left[f(x) \pm g(x)\right]$ $= L \pm M$
Constant multiple	$\lim_{x \to a} k f(x) = kL$
Product	$\lim_{x \to a} f(x) g(x)$ $= LM$
Quotient	$\lim_{x \to a} \frac{f(x)}{g(x)}$ $= \frac{L}{M}$	$M \ne 0$
Power	$\lim_{x \to a} \left[f(x)\right]^n$ $= L^n$	$n$ a positive integer
Root	$\lim_{x \to a} \sqrt{f(x)}$ $= \sqrt{L}$	$L > 0$ for even roots

Two simple limits start everything off: $\lim_{x \to a} k = k$ for a constant $k$ , and $\lim_{x \to a} x = a$ . Both are obvious from their graphs, a flat line and the line $y = x$ .

Three curves near x = 2: f approaching 3, g approaching -1, and f + g approaching 2. A panel lists: as x approaches 2, f(x) approaches 3, g(x) approaches -1, f(x) + g(x) approaches 2, f(x) times g(x) approaches -3, and 5f(x) approaches 15. — **Figure 1.** Near $x = 2$ the curve $f + g$ heads to $3 + (-1) = 2$ , the sum of the two limits. LevlPrep original diagram.

Worked example 1. Given $\lim_{x \to 4} f(x) = 3$ and $\lim_{x \to 4} g(x) = -1$ , find $\lim_{x \to 4} \frac{2f(x) - [g(x)]^2}{f(x) + g(x)}$ .

Numerator: $2(3) - (-1)^2 = 6 - 1 = 5$ , by the constant multiple, power and difference laws.

Denominator: $3 + (-1) = 2$ , which is not 0, so the quotient law applies.

So the limit is $\frac{5}{2}$ .

Direct substitution

Every polynomial is built from constants and $x$ using sums, products and constant multiples. The laws therefore give, for any polynomial $p$ , $\lim_{x \to a} p(x) = p(a)$ The same holds for a rational function $\frac{p(x)}{q(x)}$ as long as $q(a) \ne 0$ , and for roots of these where the expression is defined. Finding a limit by evaluating at $a$ is called direct substitution. It is the first thing to try on every limit.

Worked example 2. Find $\lim_{x \to 2} \frac{x^3 - 4x + 1}{x + 3}$ .

This is a rational function, and the denominator at $x = 2$ is $5 \ne 0$ . Substitute:

$\lim_{x \to 2} \frac{x^3 - 4x + 1}{x + 3} = \frac{8 - 8 + 1}{2 + 3} = \frac{1}{5}$

Why does this not contradict the idea that a limit ignores $f(a)$ ? It doesn't. For these well-behaved functions the limit and the value simply happen to agree. The laws prove they agree; the value is not being used as the definition.

When the laws are silent

The laws come with conditions, and the conditions are where exam questions live.

Denominator heading to 0, numerator not. For $\lim_{x \to 2} \frac{x + 1}{x - 2}$ , the top heads to 3 and the bottom to 0. Numbers near 3 divided by numbers near 0 are huge in size, so there is no finite limit. Writing $\frac{3}{0}$ is never a value; say it in words.
Both top and bottom heading to 0. For $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ , substitution gives a zero in both places, which tells you nothing yet. The limit might be any number, or might not exist. Topic 1.6 shows the algebra that settles it.
One of the limits does not exist. The laws need both limits to exist. If they do not, the laws do not apply, but the combination might still have a limit. Two jumps can cancel each other in a sum.

Worked example 3. $f$ jumps at $x = 1$ : it heads to 2 from the left and 0 from the right. $g$ heads to 1 from the left and 3 from the right. Find $\lim_{x \to 1} [f(x) + g(x)]$ and $\lim_{x \to 1} f(x)g(x)$ .

Neither $f$ nor $g$ has a limit at 1, so use one side at a time.

Sum: from the left $2 + 1 = 3$ ; from the right $0 + 3 = 3$ . So $\lim_{x \to 1} [f(x) + g(x)]$ $= 3$ .

Product: from the left $2 \cdot 1 = 2$ ; from the right $0 \cdot 3 = 0$ . They differ, so $\lim_{x \to 1} f(x)g(x)$ does not exist.

One-sided limits from formulas

The laws work for one-sided limits too. For a piecewise function, each one-sided limit uses the formula for that side.

Worked example 4. Let $f(x)$ $= \begin{cases} x^2 - 1 & x < 3 \\ 2x + 2 & x \ge 3 \end{cases}$ . Find $\lim_{x \to 3} f(x)$ .

Left: $\lim_{x \to 3^-} (x^2 - 1)$ $= 9 - 1$ $= 8$ . Right: $\lim_{x \to 3^+} (2x + 2)$ $= 8$ .

Both are 8, so $\lim_{x \to 3} f(x) = 8$ .

Limits of compositions

For $\lim_{x \to a} f(g(x))$ , work from the inside out. First find $b = \lim_{x \to a} g(x)$ . If $f$ is continuous at $b$ (its graph is unbroken there), then $f(g(x))$ heads to $f(b)$ : $\lim_{x \to a} f(g(x)) = f\left(\lim_{x \to a} g(x)\right) = f(b)$ For example, $\lim_{x \to 3} \sqrt{x^2 + 7}$ $= \sqrt{16}$ $= 4$ . With graphs, read the inside limit from the graph of $g$ , then read $f$ near that number from the graph of $f$ . If $f$ jumps at $b$ , check which side the inside values approach from.