Algebra for Limits: Factoring and…

Part 1 · Hook

Why this matters

The most common limit on the exam looks impossible at first: substitute, and the top and bottom are both zero. The way through is algebra you already know from precalculus: factoring, multiplying by a conjugate, combining fractions. This page sharpens exactly those moves, in the form calculus will ask for them.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Direct substitution gives $\lim_{x \to 2} (x^2 + 1) =$

$5$
$4$
$3$
it does not exist

Show the answer

A polynomial: substitute, $4 + 1 = 5$ .

Correct: $5$ :
$4$ :
$3$ :
it does not exist:

2. Which expression equals $\frac{1}{\sqrt{x}}$ ?

$x^{-1/2}$
$x^{1/2}$
$x^{-2}$
$-x^{1/2}$

Show the answer

$\sqrt{x} = x^{1/2}$ ; a reciprocal flips the sign of the exponent.

Correct: $x^{-1/2}$ :
$x^{1/2}$ :
$x^{-2}$ :
$-x^{1/2}$ :

Part 4 · See it

See it first

The graph of y = (square root of x minus 2) divided by (x minus 4) for x from 0 to 12: it starts at (0, 0.5), falls slowly, and has an open circle at (4, 1/4). A panel shows the algebra: multiplying by the conjugate gives 1 over (square root of x plus 2), which is 1/4 at x = 4. — $\frac{\sqrt{x} - 2}{x - 4}$ equals $\frac{1}{\sqrt{x} + 2}$ everywhere except $x = 4$ . Multiplying by the conjugate reveals the height of the hole, $\frac{1}{4}$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A fraction has a factor that is 0 at $x = a$ in both top and bottomfactor top and bottom so that factor is visible
Canceling the common factorgives a simpler expression equal to the original for every $x \ne a$
A square root sits next to a number, as in $\sqrt{x} - 2$ multiplying by the conjugate $\sqrt{x} + 2$ turns it into $x - 4$ , which can cancel
Fractions inside a fractiona common denominator turns the complex fraction into one fraction you can simplify

Part 6 · Key ideas

Key ideas

Factoring toolkit: common factor; difference of squares $a^2 - b^2 = (a - b)(a + b)$ ; trinomials (with the $ac$ method when the leading coefficient is not 1); cubes $a^3 \pm b^3$ $= (a \pm b)(a^2 \mp ab + b^2)$ .
Cancel factors, never terms: $\frac{x(x - 4)}{(x - 4)(x + 4)}$ $= \frac{x}{x + 4}$ , but $\frac{x^2 + 4}{x + 2}$ does not simplify.
Rationalizing: multiply top and bottom by the conjugate ( $\sqrt{A} - B$ and $\sqrt{A} + B$ form a pair). Their product is $A - B^2$ . Leave the other factor unmultiplied.
A complex fraction becomes one fraction with a common denominator; then divide.
Expand powers fully: $(a + h)^2 = a^2 + 2ah + h^2$ . The middle term is the one that matters.

Part 7 · Misconception

A common mistake

The wrong idea: $\frac{x^2 - 4x}{x^2 - 16}$ $= \frac{-4x}{-16}$ , canceling the $x^2$ terms.

What actually happens: Only factors cancel. Factor first: $\frac{x(x - 4)}{(x - 4)(x + 4)}$ $= \frac{x}{x + 4}$ . Test a number to catch the error: at $x = 1$ the original is $\frac{1}{5}$ , not $\frac{1}{4}$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. Which is the factored form of $2x^2 + 5x - 3$ ?

$(2x - 1)(x + 3)$
$(2x + 1)(x - 3)$
$(2x - 3)(x + 1)$
$(x - \tfrac{1}{2})(x + 3)$

Show the answer

Check by expanding: $(2x - 1)(x + 3)$ $= 2x^2$ $+ 6x - x - 3$ $= 2x^2$ $+ 5x - 3$ . (Method: $ac = -6$ ; the factors 6 and $-1$ add to 5, so split $5x$ as $6x - x$ and group.)

Correct: $(2x - 1)(x + 3)$ : Expands back to $2x^2 + 5x - 3$ .
$(2x + 1)(x - 3)$ : This expands to $2x^2 - 5x - 3$ : the signs are reversed.
$(2x - 3)(x + 1)$ : This expands to $2x^2 - x - 3$ .
$(x - \tfrac{1}{2})(x + 3)$ : This expands to $x^2 + \frac{5}{2}x - \frac{3}{2}$ : the leading coefficient 2 was lost.

2. Simplify $\frac{x^2 - 16}{x^2 + 2x - 8}$ for $x \ne -4$ and $x \ne 2$ .

$\frac{x - 4}{x - 2}$
$\frac{x + 4}{x + 2}$
$\frac{-16}{2x - 8}$
$\frac{x - 4}{x + 2}$

Show the answer

Factor both: $\frac{(x - 4)(x + 4)}{(x + 4)(x - 2)}$ . Cancel the common factor $x + 4$ : $\frac{x - 4}{x - 2}$ .

Correct: $\frac{x - 4}{x - 2}$ : The common factor is $x + 4$ .
$\frac{x + 4}{x + 2}$ : That factors the bottom as $(x - 4)(x + 2)$ , which expands to $x^2 - 2x - 8$ . The bottom is $(x + 4)(x - 2)$ .
$\frac{-16}{2x - 8}$ : Canceling $x^2$ terms is not allowed: only factors cancel, never terms.
$\frac{x - 4}{x + 2}$ : That factors the bottom as $(x + 4)(x + 2)$ $= x^2 + 6x + 8$ . The bottom is $(x + 4)(x - 2)$ .

3. Which fraction can be simplified by canceling a common factor?

$\frac{x^2 - 9}{x + 3}$
$\frac{x^2 + 9}{x + 3}$
$\frac{x + 9}{x + 3}$
$\frac{x^2 + 3}{x + 3}$

Show the answer

$x^2 - 9 = (x - 3)(x + 3)$ shares the factor $x + 3$ with the bottom, so it simplifies to $x - 3$ .

Correct: $\frac{x^2 - 9}{x + 3}$ : A difference of squares has the factor $x + 3$ .
$\frac{x^2 + 9}{x + 3}$ : $x^2 + 9$ does not factor over the real numbers, so nothing cancels.
$\frac{x + 9}{x + 3}$ : The 3s are inside sums, not factors; $\frac{x + 9}{x + 3}$ does not simplify.
$\frac{x^2 + 3}{x + 3}$ : $x^2 + 3$ has no factor $x + 3$ : at $x = -3$ it equals 12, not 0.

4. What is $\left(\sqrt{x + 5} - 3\right)\left(\sqrt{x + 5} + 3\right)$ ?

$x - 4$
$x + 2$
$x + 14$
$x - 4 + 6\sqrt{x + 5}$

Show the answer

A conjugate pair multiplies to a difference of squares: $(\sqrt{x + 5})^2 - 3^2$ $= x$ $+ 5 - 9$ $= x - 4$ . The middle terms cancel.

Correct: $x - 4$ : $(x + 5) - 9$ .
$x + 2$ : That subtracts 3 instead of $3^2 = 9$ .
$x + 14$ : The 9 is subtracted, not added: $(a - b)(a + b) = a^2 - b^2$ .
$x - 4 + 6\sqrt{x + 5}$ : The cross terms $+3\sqrt{x + 5}$ and $-3\sqrt{x + 5}$ cancel.

5. Simplify the complex fraction $\frac{\frac{1}{x + 2} - \frac{1}{2}}{x}$ for $x \ne 0$ and $x \ne -2$ .

$\frac{-1}{2(x + 2)}$
$\frac{1}{2(x + 2)}$
$\frac{-x}{2(x + 2)}$
$\frac{-1}{x + 2}$

Show the answer

Common denominator on top: $\frac{2 - (x + 2)}{2(x + 2)}$ $= \frac{-x}{2(x + 2)}$ . Dividing by $x$ cancels the $x$ : $\frac{-1}{2(x + 2)}$ .

Correct: $\frac{-1}{2(x + 2)}$ : Top: $\frac{-x}{2(x + 2)}$ ; divide by $x$ .
$\frac{1}{2(x + 2)}$ : $2 - (x + 2) = -x$ : the minus sign must be distributed.
$\frac{-x}{2(x + 2)}$ : That is only the numerator; it must still be divided by $x$ .
$\frac{-1}{x + 2}$ : The common denominator is $2(x + 2)$ ; the 2 cannot be dropped.

6. Simplify $\frac{(3 + h)^2 - 9}{h}$ for $h \ne 0$ .

$6 + h$
$h$
$6 + h^2$
$9 + h$

Show the answer

Expand: $(3 + h)^2 - 9$ $= 9$ $+ 6h$ $+ h^2 - 9$ $= 6h$ $+ h^2$ $= h(6 + h)$ . Divide by $h$ : $6 + h$ .

Correct: $6 + h$ : $\frac{h(6 + h)}{h}$ .
$h$ : That uses $(3 + h)^2 = 9 + h^2$ , missing the middle term $6h$ .
$6 + h^2$ : Dividing $h^2$ by $h$ gives $h$ , not $h^2$ .
$9 + h$ : The 9s cancel; nothing leaves a 9.

7. A student simplifies: $\frac{x^2 - 4x}{x^2 - 16}$ $= \frac{-4x}{-16}$ $= \frac{x}{4}$ . What went wrong?

The $x^2$ terms were canceled, but only factors can cancel; correctly, $\frac{x(x - 4)}{(x - 4)(x + 4)}$ $= \frac{x}{x + 4}$ .
Nothing: the result $\frac{x}{4}$ is correct.
The student should have canceled $x$ instead: $\frac{x - 4}{x - 16}$ .
The negatives should not cancel, so the answer is $-\frac{x}{4}$ .

Show the answer

Canceling works on factors (things multiplied), not on terms (things added). Factor first: $x^2 - 4x = x(x - 4)$ and $x^2 - 16 = (x - 4)(x + 4)$ . Then cancel $x - 4$ .

Correct: The $x^2$ terms were canceled, but only factors can cancel; correctly, $\frac{x(x - 4)}{(x - 4)(x + 4)}$ $= \frac{x}{x + 4}$ .: Factor, then cancel the common factor $x - 4$ .
Nothing: the result $\frac{x}{4}$ is correct.: Test $x = 1$ : $\frac{1 - 4}{1 - 16} = \frac{1}{5}$ , but $\frac{1}{4} \ne \frac{1}{5}$ .
The student should have canceled $x$ instead: $\frac{x - 4}{x - 16}$ .: $x$ is a factor of the top only, not of $x^2 - 16$ .
The negatives should not cancel, so the answer is $-\frac{x}{4}$ .: The real error is canceling terms; the signs are not the issue.

Part 9 · Summary

Summary

Limits that give zero over zero are settled with precalculus algebra: factor and cancel a common factor, multiply by a conjugate to clear a square root, or combine a complex fraction into one. Cancel factors, never terms, and expand powers in full. The simplified expression agrees with the original everywhere except the troublesome point.

Part 10 · Up next

What comes next

Part 11 · Connections