Selecting Procedures for Determining Limits (AP Calculus AB 1.7)

The limits $\lim\limits_{x \to 1}\dfrac{x^2 - 1}{x - 1}$ , $\lim\limits_{x \to 1}\dfrac{x^2 + 1}{x - 1}$ and $\lim\limits_{x \to 1}\dfrac{|x - 1|}{x - 1}$ have the same denominator and need three different responses: the first is 2 after factoring, the second has no finite limit, and the third has different one-sided limits, $-1$ and $1$ . No limit tells you its method. You find it by substituting first and then reading two things: the form substitution produced, and the structure of the expression.

Substitute first, then read the result

Substituting $x = a$ is always the first move, because it either finishes the problem or tells you what kind of problem you have.

Key ideaWhat substitution tells you

A real number. If the function is built from polynomials, roots, trig, exponential and log functions, and $a$ is inside its domain and not a point where a piecewise formula changes, that number is the limit (topic 1.5).
A nonzero number over 0. There is no finite limit, and no algebra will change that. Check the sign on each side to decide between $\infty$ and $-\infty$ (topic 1.14).
Zero over zero. The form $\frac{0}{0}$ is indeterminate: rewrite the expression (topic 1.6) and substitute again.
Substitution doesn't apply. At a point where the formula changes (a piecewise function or an absolute value), or where the function is defined on only one side, work with one-sided limits. If a factor like $\sin\frac{1}{x}$ has no value or limit at $a$ , try the squeeze theorem (topic 1.8).

After any rewrite, read the result of substituting again. A $\frac{0}{0}$ can become a number, another $\frac{0}{0}$ (rewrite again), or a nonzero number over 0 (no finite limit).

Matching a $\frac{0}{0}$ to its rewrite

The structure of the expression tells you which rewrite will expose a common factor.

What you see	What to do	Example
Polynomials on top and bottom	Factor and cancel	$\dfrac{x^2 - 1}{x - 1} = x + 1$
A square root in a sum or difference	Multiply by the conjugate	$\dfrac{\sqrt{x + 1} - 1}{x}$
Fractions inside a fraction	Combine over a common denominator	$\dfrac{\frac{1}{x} - \frac{1}{2}}{x - 2}$
A power that isn't expanded	Expand, then cancel	$\dfrac{(2 + h)^3 - 8}{h}$
Trig functions with a factorable identity	Rewrite with the identity	$\dfrac{\sin 2x}{\cos x}$ near $\frac{\pi}{2}$
$\sin(\text{something} \to 0)$ over a power of $x$	Arrange $\frac{\sin\theta}{\theta}$ and use its limit, 1	$\dfrac{\sin 4x}{x}$

Sometimes two methods work. In $\dfrac{x - 9}{\sqrt{x} - 3}$ you can multiply by the conjugate, or you can see $x - 9$ as a difference of squares, $(\sqrt{x} - 3)(\sqrt{x} + 3)$ , and cancel directly. Either is correct; the factoring is shorter.

The last row of the table belongs to topic 1.8, which proves $\lim\limits_{\theta \to 0}\frac{\sin\theta}{\theta} = 1$ . To use it, the angle inside the sine and the quantity in the denominator must match and both approach 0: $\frac{\sin 4x}{x} = 4\cdot\frac{\sin 4x}{4x} \to 4 \cdot 1 = 4$ .

When the formula changes: absolute values and pieces

An absolute value is a piecewise function in disguise: $|x - a| = x - a$ when $x \geq a$ , and $|x - a| = -(x - a)$ when $x < a$ . At $x = a$ the formula switches, so split the limit into two sides and remove the absolute value on each:

\lim_{x \to 3^+}\frac{x^2 - 9}{|x - 3|} = \lim_{x \to 3^+}\frac{(x - 3)(x + 3)}{x - 3} = 6, \qquad \lim_{x \to 3^-}\frac{x^2 - 9}{|x - 3|} = \lim_{x \to 3^-}\frac{(x - 3)(x + 3)}{-(x - 3)} = -6.

The one-sided limits differ, so $\lim\limits_{x \to 3}\frac{x^2 - 9}{|x - 3|}$ does not exist. A piecewise function works the same way, with the formulas given to you: each one-sided limit uses the formula for its own side, and each of those may need its own method.

When no rewrite works

Some limits can't be fixed by algebra, and recognizing them saves time.

A bounded factor that oscillates, times a factor that approaches 0, such as $x\cos\frac{1}{x}$ at $0$ . The oscillating factor has no limit, so the product law fails, but the squeeze theorem (topic 1.8) gives 0.
Only a graph or a table is given. Read or estimate the one-sided limits from it (topics 1.3 and 1.4); there is no formula to manipulate.
Unit 4 adds a tool. L'Hospital's rule (topic 4.7) settles many $\frac{0}{0}$ limits using derivatives. Until then, the methods above are your tools, and they stay the faster route for most rational and radical expressions.

Worked examples

Worked exampleExample 1: classify, then evaluate

Find each limit, or explain why it does not exist.

(a) $\lim\limits_{x \to 2}\dfrac{x^2 + 1}{x + 3}$ (b) $\lim\limits_{x \to -1}\dfrac{x^2 - 1}{x^2 + 2x + 1}$ (c) $\lim\limits_{x \to 9}\dfrac{x - 9}{\sqrt{x} - 3}$ (d) $\lim\limits_{x \to 0}\dfrac{x + |x|}{x}$

Show solution

(a) Substitution gives $\frac{5}{5} = 1$ . The denominator isn't 0 and the function is rational, so the limit is 1.

(b) Substitution gives $\frac{1 - 1}{1 - 2 + 1} = \frac{0}{0}$ , and both are polynomials, so factor:

\frac{(x - 1)(x + 1)}{(x + 1)^2} = \frac{x - 1}{x + 1}\quad\text{for } x \neq -1.

Substituting again gives $\frac{-2}{0}$ , a nonzero number over 0, so there is no finite limit. Checking signs: for $x$ just left of $-1$ , $x + 1$ is a small negative number, so $\frac{-2}{\text{small negative}} \to \infty$ ; just right of $-1$ the quotient goes to $-\infty$ . The limit does not exist.

\lim_{x \to 9}\left(\sqrt{x} + 3\right) = 6.

(d) The absolute value switches formula at 0, so split. For $x > 0$ , $|x| = x$ and the expression is $\frac{2x}{x} = 2$ . For $x < 0$ , $|x| = -x$ and the expression is $\frac{0}{x} = 0$ . The right-hand limit is 2 and the left-hand limit is 0, so the limit does not exist.

Worked exampleExample 2: two trig limits that need different tools

Find (a) $\lim\limits_{x \to 0}\dfrac{\sin(x^2)}{x}$ and (b) $\lim\limits_{x \to 0}\dfrac{1 - \cos x}{x\sin x}$ .

Show solution

Both are $\frac{0}{0}$ at $x = 0$ , and neither has a factor you can cancel directly.

(a) The sine's argument is $x^2$ , so the known limit needs $x^2$ in the denominator. Multiply and divide by $x$ :

\frac{\sin(x^2)}{x} = x\cdot\frac{\sin(x^2)}{x^2}.

As $x \to 0$ , $\theta = x^2 \to 0$ , so $\frac{\sin(x^2)}{x^2} \to 1$ , and the product approaches $0 \cdot 1 = 0$ .

(b) The $1 - \cos x$ suggests the identity $1 - \cos^2 x = \sin^2 x$ , which needs a factor of $1 + \cos x$ . Multiply top and bottom by it:

\frac{(1 - \cos x)(1 + \cos x)}{x\sin x\,(1 + \cos x)} = \frac{\sin^2 x}{x\sin x\,(1 + \cos x)} = \frac{\sin x}{x}\cdot\frac{1}{1 + \cos x}.

The first factor approaches 1 and the second approaches $\frac{1}{2}$ , so the limit is $\frac{1}{2}$ .

Worked exampleExample 3: one function, three procedures (exam style, no calculator)

Let $f$ be the function defined by

f(x) = \begin{cases} \dfrac{x^2 - 9}{x^2 - 2x - 3}, & x < 3 \\[2ex] \dfrac{6\left(\sqrt{x + 1} - 2\right)}{x - 3}, & x > 3. \end{cases}

(a) Find $\lim\limits_{x \to 0} f(x)$ .

(b) Find $\lim\limits_{x \to -1^-} f(x)$ and $\lim\limits_{x \to -1^+} f(x)$ .

Show solution

For $x < 3$ , factor once and reuse it: $\dfrac{x^2 - 9}{x^2 - 2x - 3} = \dfrac{(x - 3)(x + 3)}{(x - 3)(x + 1)} = \dfrac{x + 3}{x + 1}$ for $x \neq 3$ , $x \neq -1$ .

(a) Near 0 the first formula applies, and substitution works: $\dfrac{0 - 9}{0 - 0 - 3} = 3$ .

(b) Near $-1$ the first formula applies, and $\dfrac{x + 3}{x + 1}$ gives $\frac{2}{0}$ at $x = -1$ , a nonzero number over 0. The numerator is near 2. Just left of $-1$ , $x + 1$ is a small negative number, so $\lim\limits_{x \to -1^-} f(x) = -\infty$ ; just right, it is a small positive number, so $\lim\limits_{x \to -1^+} f(x) = \infty$ .

From the left, $\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^-}\dfrac{x + 3}{x + 1} = \dfrac{6}{4} = \dfrac{3}{2}$ .

From the right, substitution gives $\frac{0}{0}$ and there is a root in a difference, so multiply by the conjugate:

\frac{6\left(\sqrt{x + 1} - 2\right)}{x - 3}\cdot\frac{\sqrt{x + 1} + 2}{\sqrt{x + 1} + 2} = \frac{6\,(x + 1 - 4)}{(x - 3)\left(\sqrt{x + 1} + 2\right)} = \frac{6}{\sqrt{x + 1} + 2},

so $\lim\limits_{x \to 3^+} f(x) = \dfrac{6}{2 + 2} = \dfrac{3}{2}$ .

Yes, the limit exists: $\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^+} f(x) = \frac{3}{2}$ , so $\lim\limits_{x \to 3} f(x) = \frac{3}{2}$ . This is true even though $f(3)$ is not defined.

Where students lose points

On the exam

Limit questions give you the expression and ask for the value, without naming a method, so the recognition step in this topic is part of every one of them. When a free-response part asks whether a limit exists, especially for a piecewise function like Example 3, the justification is both one-sided limits, each shown, and a statement that they are equal (or that they differ).

Check yourself

Question 1
Substituting $x = 2$ into a rational function $r(x)$ gives $\dfrac{3}{0}$ . Which statement is true?
1. A
  $\lim\limits_{x \to 2} r(x) = 0$
2. B
  $\lim\limits_{x \to 2} r(x) = 3$
3. C
  Factoring and canceling will reveal $\lim\limits_{x \to 2} r(x)$ .
4. D
  $\lim\limits_{x \to 2} r(x)$ is not a real number.
Show answer
D
A numerator near 3 divided by a denominator near 0 has unbounded size, so there is no finite limit. Choice C is the tempting one, but the denominator's factor $x - 2$ isn't in the numerator (the numerator is 3 there), so nothing cancels.
Question 2
Find $\lim\limits_{x \to 0}\dfrac{\sin 3x}{x^2 + 2x}$ .
Show answer
$\dfrac{3}{2}$
Factor the denominator as $x(x + 2)$ and rearrange: $3\cdot\dfrac{\sin 3x}{3x}\cdot\dfrac{1}{x + 2}$ . The middle factor approaches 1 (topic 1.8) and the last approaches $\frac{1}{2}$ , giving $\frac{3}{2}$ .
Question 3
What is $\lim\limits_{x \to 2^-}\dfrac{|x - 2|}{x^2 - 4}$ ?
1. A
  $-\dfrac{1}{4}$
2. B
  $\dfrac{1}{4}$
3. C
  $0$
4. D
  $-\infty$
Show answer
A
For $x < 2$ , $|x - 2| = -(x - 2)$ , so the expression is $\dfrac{-(x - 2)}{(x - 2)(x + 2)} = -\dfrac{1}{x + 2}$ , which approaches $-\frac{1}{4}$ . Choice B drops the sign the absolute value introduces on the left side.
Question 4
For which function does substituting $x = 0$ give $\lim\limits_{x \to 0} f(x)$ directly?
1. A
  $f(x) = \dfrac{\cos x}{1 + x}$
2. B
  $f(x) = \dfrac{\sin x}{x}$
3. C
  $f(x) = \dfrac{|x|}{x}$
4. D
  $f(x) = x\sin\dfrac{1}{x}$
Show answer
A
In A the denominator is $1 \neq 0$ at $x = 0$ and every piece is continuous, so the limit is $\frac{1}{1} = 1$ . B is $\frac{0}{0}$ (its limit, 1, takes topic 1.8), C changes formula at 0, and D is not defined at 0 and needs the squeeze theorem.
Question 5
Find $\lim\limits_{h \to 0}\dfrac{\frac{1}{(1 + h)^2} - 1}{h}$ .
Show answer
$-2$
Combine the top over the common denominator $(1 + h)^2$ : the expression becomes $\dfrac{1 - (1 + h)^2}{h(1 + h)^2} = \dfrac{-2h - h^2}{h(1 + h)^2} = \dfrac{-2 - h}{(1 + h)^2}$ for $h \neq 0$ , which approaches $-2$ .

Selecting Procedures for Determining Limits

Substitute first, then read the result

Matching a 00\frac{0}{0}00​ to its rewrite

When the formula changes: absolute values and pieces

When no rewrite works

Worked examples

Where students lose points

On the exam

Check yourself

Go deeper

Matching a $\frac{0}{0}$ to its rewrite