Math Core

Tables and graphs are great for building intuition, but they only ever give estimates. The limit laws let you compute limits exactly with algebra, and they explain why most limits can be found just by plugging in, and what to do when plugging in fails.

The limit laws

Limits behave nicely with arithmetic. If two functions approach known values, their sum approaches the sum of those values, their product approaches the product, and so on.

Limit laws

Suppose $\displaystyle \lim_{x \to a} f(x) = L$ and $\displaystyle \lim_{x \to a} g(x) = M$ , and $c$ is a constant. Then:

Law	Statement
Sum and difference	$\displaystyle \lim_{x \to a} [f(x) \pm g(x)] = L \pm M$
Constant multiple	$\displaystyle \lim_{x \to a} c\,f(x) = cL$
Product	$\displaystyle \lim_{x \to a} f(x)\,g(x) = LM$
Quotient	$\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$ , provided $M \ne 0$
Power	$\displaystyle \lim_{x \to a} [f(x)]^n = L^n$ for a positive integer $n$
Root	$\displaystyle \lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}$ (for even $n$ , need $L > 0$ )

Two very simple limits get everything started: $\displaystyle \lim_{x \to a} c = c$ (a constant stays put) and $\displaystyle \lim_{x \to a} x = a$ (the input approaches $a$ by definition). Combine them with the laws and you can take the limit of any polynomial.

Worked example: Using the laws with given limits

Suppose $\displaystyle \lim_{x \to 1} f(x) = 5$ and $\displaystyle \lim_{x \to 1} g(x) = -2$ . Find $\displaystyle \lim_{x \to 1} \frac{3f(x) + g(x)^2}{f(x) - g(x)}$ .

Solution. The denominator's limit is $5 - (-2) = 7$ , which is not $0$ , so the quotient law applies.

\lim_{x \to 1} \frac{3f(x) + g(x)^2}{f(x) - g(x)} = \frac{3(5) + (-2)^2}{5 - (-2)} = \frac{15 + 4}{7} = \frac{19}{7}.

Direct substitution

Apply the laws to a polynomial like $p(x) = 2x^3 - x + 4$ and you find that $\displaystyle \lim_{x \to a} p(x) = 2a^3 - a + 4 = p(a)$ . The limit is just the value. The same happens for rational functions, as long as the denominator isn't zero at $a$ .

Direct substitution

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

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

For example, $\displaystyle \lim_{x \to 2} (3x^2 - 5x + 1) = 3(4) - 10 + 1 = 3$ , and $\displaystyle \lim_{x \to -1} \frac{x + 5}{x^2 + 1} = \frac{4}{2} = 2$ . Square roots, exponentials, logarithms and trig functions also allow direct substitution at any point inside their domains.

So the first move for any limit is always the same: try plugging in. If you get a real number, you're done.

When substitution gives $0/0$

The interesting limits are the ones where substitution breaks. The most important case is $\dfrac{0}{0}$ , as in

\lim_{x \to 3} \frac{x^2 - x - 6}{x - 3}.

The result $\dfrac{0}{0}$ is called an indeterminate form. It doesn't tell you the answer; it tells you the numerator and denominator are both shrinking to $0$ , and the limit depends on how fast each one shrinks. Your job is to rewrite the expression so the troublesome factor disappears. Three algebra tools cover most cases:

Factor and cancel. A zero at $x = a$ for both a polynomial numerator and denominator means both share a factor of $(x - a)$ .
Rationalize. If a square root is involved, multiply the top and bottom by the conjugate.
Combine fractions. If there are fractions inside a fraction, rewrite with a common denominator.

Cancelling is legal because a limit as $x \to a$ never uses $x = a$ itself. For every $x \ne a$ the original expression and the simplified one are equal, so they have the same limit.

Common mistake

$\dfrac{0}{0}$ does not mean the limit is $0$ , or $1$ , or that it doesn't exist. It means "more work needed." The same form can hide any answer at all.

Worked example: Factor and cancel

Find $\displaystyle \lim_{x \to 3} \frac{x^2 - x - 6}{x - 3}$ .

Solution. Substitution gives $\dfrac{9 - 3 - 6}{0} = \dfrac{0}{0}$ . Factor the numerator:

\lim_{x \to 3} \frac{(x - 3)(x + 2)}{x - 3} = \lim_{x \to 3} (x + 2) = 5.

The graph of the original function is the line $y = x + 2$ with a hole at $(3, 5)$ .

y = (x² − x − 6)/(x − 3): the line y = x + 2 with a hole at (3, 5).Open in grapher →

Worked example: Rationalize

Find $\displaystyle \lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$ .

Solution. Substitution gives $\dfrac{3 - 3}{0} = \dfrac{0}{0}$ . Multiply the top and bottom by the conjugate $\sqrt{x + 9} + 3$ :

\begin{aligned} \frac{\sqrt{x + 9} - 3}{x} \cdot \frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3} &= \frac{(x + 9) - 9}{x\left(\sqrt{x + 9} + 3\right)} \\ &= \frac{x}{x\left(\sqrt{x + 9} + 3\right)} = \frac{1}{\sqrt{x + 9} + 3}. \end{aligned}

Now substitute: $\dfrac{1}{\sqrt{9} + 3} = \dfrac{1}{6}$ .

Worked example: Combine fractions

Find $\displaystyle \lim_{x \to 4} \frac{\frac{1}{x} - \frac{1}{4}}{x - 4}$ .

Solution. Substitution gives $\dfrac{0}{0}$ . Combine the top over the common denominator $4x$ :

\frac{1}{x} - \frac{1}{4} = \frac{4 - x}{4x} = \frac{-(x - 4)}{4x}.

Dividing by $x - 4$ cancels that factor, leaving $\dfrac{-1}{4x}$ . Substitute: $\dfrac{-1}{4(4)} = -\dfrac{1}{16}$ .

When substitution gives (nonzero) $/0$

If the numerator approaches a nonzero number and the denominator approaches $0$ , the fraction's size blows up. For instance, as $x \to 2$ , $\dfrac{x + 1}{x - 2}$ has numerator near $3$ and denominator near $0$ . From the right the values are huge and positive; from the left they are huge and negative. The graph has a vertical asymptote at $x = 2$ and the limit does not exist. No algebra trick will change that, because there is no common factor to cancel.

Tip

Summary of the decision: substitute first. A real number means you're done. $\dfrac{0}{0}$ means simplify and try again. (Nonzero) $/0$ means a vertical asymptote, and the two-sided limit does not exist.

Practice

Practice 1

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