Math · Calculus I · Worked example

Find a limit by factoring and canceling

Find the limit of $(x^{2} - 4) / (x - 2)$ as $x$ approaches $2$ .

lim_{x \to 2} \frac{x^{2} - 4}{x - 2}

Find where the expression is defined

The denominator $x - 2$ is $0$ at $x = 2$ , so the expression is undefined there. The limit asks about inputs near $2$ , not at it.

f (x) = \frac{x^{2} - 4}{x - 2}, x \neq 2

Try substitution

Both the numerator and the denominator become $0$ . $0 / 0$ is not a number, so substitution has not answered the question. It says the expression needs rewriting.

\frac{2^{2} - 4}{2 - 2} = \frac{0}{0}

Factor the numerator

$x^{2} - 4$ is a difference of squares, $a^{2} - b^{2} = (a - b) (a + b)$ with $a = x$ and $b = 2$ . The factor $x - 2$ now appears on top and on the bottom.

x^{2} - 4 = (x - 2) (x + 2)

Cancel, keeping $x \neq 2$

For every input other than $2$ , $x - 2$ is not $0$ , and dividing by it is allowed. Keep $x \neq 2$ beside the result, because the original expression is still undefined at $2$ .

\frac{(x - 2) (x + 2)}{x - 2} = x + 2 (x \neq 2)

Take the limit of the simpler form

Near $2$ the quotient equals $x + 2$ , so it has the same limit. $x + 2$ is a polynomial, and substitution gives $2 + 2 = 4$ .

lim_{x \to 2} \frac{x^{2} - 4}{x - 2} = lim_{x \to 2} (x + 2) = 4

Keep the limit and the value apart

The limit is $4$ , and the function is still undefined at $2$ . Its graph is the line $y = x + 2$ with a hole at $(2, 4)$ .

The limit and the value answer different questions
Question	Answer
Limit from the left	$4$
Limit from the right	$4$
Two-sided limit	$4$
Value of $f$ at $2$	undefined

Result

The limit is $4$ . The function itself is undefined at $x = 2$ , where its graph has a hole at $(2, 4)$ .

lim_{x \to 2} \frac{x^{2} - 4}{x - 2} = 4

Your turn

1. One like the example

Find the limit of $(x^{2} - 9) / (x - 3)$ as $x$ approaches $3$ . Is the function defined at $3$ ?

Show the answer and explanation

The limit is $6$ . The function is undefined at $3$ .

Substitution gives $0 / 0$ . Factor $x^{2} - 9$ as $(x - 3) (x + 3)$ and cancel $x - 3$ , which is allowed for $x \neq 3$ . Then $x + 3$ approaches $3 + 3 = 6$ . The original denominator is still $0$ at $x = 3$ , so $f (3)$ does not exist.

\begin{array}{l} lim_{x \to 3} \frac{x^{2} - 9}{x - 3} \\ lim_{x \to 3} (x + 3) \\ 6 \end{array}

2. A variation

Find the limit of $(x^{2} + 3 x + 2) / (x^{2} - 1)$ as $x$ approaches $- 1$ . This time both the top and the bottom need factoring.

Show the answer and explanation

The limit is $- 1 / 2$ .

At $x = - 1$ the top is $1 - 3 + 2 = 0$ and the bottom is $1 - 1 = 0$ . Factor both, $(x + 1) (x + 2)$ over $(x + 1) (x - 1)$ , and cancel $x + 1$ . Substituting into $(x + 2) / (x - 1)$ gives $1 / (- 2) = - 1 / 2$ .

\begin{array}{l} lim_{x \to - 1} \frac{x^{2} + 3 x + 2}{x^{2} - 1} \\ lim_{x \to - 1} \frac{x + 2}{x - 1} \\ - \frac{1}{2} \end{array}

3. Somewhere new

A ball thrown upward is $s (t) = 20 t - 5 t^{2}$ meters high after $t$ seconds. Its average velocity between $1 second$ and $t$ seconds is $(s (t) - 15) / (t - 1)$ . What does that average approach as $t$ approaches $1$ ?

Show the answer and explanation

$10$ meters per second, the ball’s velocity at $t = 1$ .

$s (1) = 20 - 5 = 15$ , so the average is $(20 t - 5 t^{2} - 15) / (t - 1)$ , which gives $0 / 0$ at $t = 1$ . The top is $- 5 (t^{2} - 4 t + 3)$ $= - 5 (t - 1) (t - 3)$ . Canceling $t - 1$ leaves $- 5 (t - 3) = 15 - 5 t$ , which approaches $15 - 5 = 10$ .

\begin{array}{l} lim_{t \to 1} \frac{20 t - 5 t^{2} - 15}{t - 1} \\ lim_{t \to 1} (15 - 5 t) \\ 10 \end{array}

Keep exploring

Open Explore in Limits and narrow the window: the left and right tables close in on $4$ from below and above, while the value at $2$ stays undefined.

Return to the concept →

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After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 5 key terms Explore in Limits Inspect nearby values in Graph Check the limit in Math Test ε and δ at the hole in The ε–δ Limit Open worked example on a board Calculus formulas

Your existing work stays on this device. Examples open as editable copies.

Find where the expression is defined

Try substitution

Factor the numerator

Cancel, keeping x≠2x=2

Take the limit of the simpler form

Keep the limit and the value apart

Result

Your turn

1. One like the example

The limit is 66. The function is undefined at 33.

2. A variation

The limit is −1⁢/2−1/2.

3. Somewhere new

1010 meters per second, the ball’s velocity at t=1t=1.

Keep exploring

Study it, then try it

Cancel, keeping $x \neq 2$

The limit is $6$ . The function is undefined at $3$ .

The limit is $- 1 / 2$ .

$10$ meters per second, the ball’s velocity at $t = 1$ .