2.3 The Limit Laws

Big idea

Tables and graphs suggest limits; they do not establish them. A table can only ever show finitely many inputs, and a limit is a claim about all of them near a point. This section replaces the evidence with algebra by proving, once, that limits respect the arithmetic you already use.

The laws all say the same kind of thing. If two functions have limits at a point, then their sum, difference, product and quotient have limits there too, and those limits are the sum, difference, product and quotient of the originals. Build up from the two simplest cases, a constant function and the function $x$ itself, and you can evaluate the limit of any polynomial by substitution, and any rational function by substitution wherever the denominator is not zero. That single consequence handles most limits you will ever meet, and it does so without a table.

What the laws do not handle is exactly the interesting case. Every derivative in this course is a limit of a quotient in which numerator and denominator both go to zero, and the quotient law explicitly requires a nonzero denominator limit, so it has nothing to say. The form $0/0$ is called indeterminate because the answer is not determined by the form: the limit can be any number at all, or fail to exist, depending on how the two parts vanish against each other.

The strategy for an indeterminate form is always the same. Since a limit ignores the point itself, you are free to replace the function by any expression that agrees with it at every other nearby input, and then apply the laws to the replacement. Factoring and cancelling is one way to produce such a replacement. Multiplying by a conjugate is another. Clearing a compound fraction is a third. The algebra differs; the logic does not.

Decoder

If two functions have limits at a point, the limit of their sum, difference, product or quotient is the corresponding combination of those limits, provided that in the quotient case the limit of the denominator is not zero.

The words carrying the weight are “if two functions have limits.” Each law is conditional. Applying it to a piece whose limit does not exist proves nothing, and a limit that fails to exist can still combine with another into a perfectly good one, which is why the laws can never be read backwards.

The proviso on the quotient is the entire reason this chapter is long. Every time the denominator limit is nonzero you are done in one line. Every time it is zero you are starting, not finishing, and what you do next depends on the numerator: a nonzero numerator limit gives an infinite limit or a one-sided pair, while a zero numerator limit gives the indeterminate form and demands rewriting.

The classic mistake is treating $0/0$ as an answer, usually as $0$ or as $1$. It is neither, and a pair of examples settles it: $\lim_{x\to 0} \frac{x^2}{x} = 0$ while $\lim_{x\to 0} \frac{x}{x^2}$ does not exist, and both have the same form. The form tells you which tool to reach for, nothing more.

Definitions and results

The two basic limits. For any real $a$ and any constant $c$, $\lim_{x \to a} c = c$ and $\lim_{x \to a} x = a$. Every other limit law is built on these two.

Sum, difference and constant multiple. If $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} g(x) = M$, then

$$ \lim_{x\to a}\big(f(x) \pm g(x)\big) = L \pm M, \qquad \lim_{x\to a} c f(x) = cL $$

Product and quotient. Under the same hypotheses, $\lim_{x\to a} f(x)g(x) = LM$, and if $M \ne 0$,

$$ \lim_{x\to a} \frac{f(x)}{g(x)} = \frac{L}{M} $$

Powers and roots. For a positive integer $n$, $\lim_{x\to a} \big(f(x)\big)^n = L^n$, and $\lim_{x\to a} \sqrt[n]{f(x)} = \sqrt[n]{L}$ whenever the root is defined, which for even $n$ requires $L > 0$ or a one-sided reading at $L = 0$.

Polynomials and rational functions. Applying the laws to a polynomial $p$ gives $\lim_{x\to a} p(x) = p(a)$. For a rational function $p/q$, if $q(a) \ne 0$ then the limit is $p(a)/q(a)$. Substitution is a theorem here, not a definition.

Replacement principle. If $f(x) = g(x)$ for all $x$ in some open interval around $a$ except possibly at $a$ itself, then $\lim_{x\to a} f(x) = \lim_{x\to a} g(x)$, meaning one exists exactly when the other does and they agree. Every cancellation in this section is an application of this principle.

One-sided laws. All of the laws above hold verbatim for one-sided limits. Since a two-sided limit exists exactly when the two sides agree, a problem with an absolute value or a piecewise definition is handled by splitting into two one-sided calculations and comparing.

The squeeze theorem. If $g(x) \le f(x) \le h(x)$ for all $x$ near $a$ except possibly at $a$, and $\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L$, then $\lim_{x\to a} f(x) = L$. For instance $-x^2 \le x^2\cos(1/x) \le x^2$ for $x \ne 0$, and both bounds go to $0$, so $\lim_{x\to 0} x^2\cos(1/x) = 0$ even though the cosine factor never settles.

Worked examples

Substitution, justified

Evaluate $\displaystyle\lim_{x\to 3} \frac{x^2 + 1}{x + 2}$.

The numerator is a polynomial, so its limit is $3^2 + 1 = 10$. The denominator is a polynomial with limit $3 + 2 = 5$, which is not zero, so the quotient law applies:

$$ \lim_{x\to 3} \frac{x^2+1}{x+2} = \frac{10}{5} = 2 $$

The work is one line, but the sentence that matters is the one checking that the denominator limit is nonzero. That check is what separates a legal substitution from a wrong answer.

Cancelling a common factor

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

Substitution gives $0/0$, so the quotient law is unavailable. The vanishing of the numerator at $x = -3$ guarantees that $x + 3$ divides it, and indeed $x^2 + x - 6 = (x+3)(x-2)$. For every $x \ne -3$,

$$ \frac{x^2+x-6}{x+3} = \frac{(x+3)(x-2)}{x+3} = x - 2 $$

The two expressions agree everywhere except at the one deleted point, so the replacement principle lets you swap them:

$$ \lim_{x\to -3} \frac{x^2+x-6}{x+3} = \lim_{x\to -3} (x-2) = -5 $$

Multiplying by the conjugate

Evaluate $\displaystyle\lim_{x\to 7} \frac{\sqrt{x+2} - 3}{x - 7}$.

Substitution gives $0/0$ again, and there is no factor to pull out of a difference of a root and a number. Multiply top and bottom by the conjugate $\sqrt{x+2} + 3$, which is nonzero near $x = 7$, so the manipulation is legal:

$$ \frac{\sqrt{x+2}-3}{x-7} \cdot \frac{\sqrt{x+2}+3}{\sqrt{x+2}+3} = \frac{(x+2) - 9}{(x-7)\big(\sqrt{x+2}+3\big)} = \frac{x-7}{(x-7)\big(\sqrt{x+2}+3\big)} $$

Cancelling the common factor leaves $\dfrac{1}{\sqrt{x+2}+3}$, whose denominator has limit $\sqrt{9} + 3 = 6$, not zero. So the limit is $\dfrac{1}{6}$.

Check against a number. At $x = 7.06$ the numerator is $\sqrt{9.06} - 3 \approx 0.009983$ and the denominator is $0.06$, giving about $0.1664$, close to $1/6 \approx 0.1667$.

Clearing a compound fraction

Evaluate $\displaystyle\lim_{h\to 0} \frac{\frac{1}{3+h} - \frac{1}{3}}{h}$.

This is the difference quotient for $1/x$ at $x = 3$, so the $0/0$ form is expected. Combine the two fractions on top over the common denominator $3(3+h)$:

$$ \frac{1}{3+h} - \frac{1}{3} = \frac{3 - (3+h)}{3(3+h)} = \frac{-h}{3(3+h)} $$

Dividing by $h$ means multiplying by $1/h$, and since $h \ne 0$ throughout the approach, the factors cancel:

$$ \frac{-h}{3(3+h)} \cdot \frac{1}{h} = \frac{-1}{3(3+h)} \longrightarrow \frac{-1}{9} $$

Check the sign before accepting it. The function $1/x$ decreases on the positive axis, so its rate of change at $x = 3$ must be negative, and $-1/9$ is.

Practice

Start with the laws in their direct form. These drills ask for limits you can evaluate by substitution once you have confirmed that the denominator does not vanish, plus limits of sums, products and roots.

Then the indeterminate quotients that factor. Find the common factor, cancel it, and evaluate what remains.

Last, the ones with radicals. Multiply by the conjugate of whichever part carries the root, simplify the difference of squares, and cancel.

Quiz

Six items on applying the limit laws, resolving a zero over zero form by factoring, and rationalizing to clear a radical.