Algebraic Techniques for Limits
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 2.3: “The Limit Laws” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Key idea
When direct substitution gives $\frac{0}{0}$, both numerator and denominator are zero at the point. That shared zero means there is a common factor -- something in the numerator and denominator that cancels. Once you cancel it, the troublesome point is no longer in the denominator, and direct substitution works on what remains.
The three algebraic moves in this lesson are all ways to expose and cancel that shared factor: factor and cancel, rationalize by multiplying by a conjugate, or combine fractions over a common denominator. The algebra changes the form of the expression without changing its limit, because you are never dividing by zero during the manipulation -- you do the algebra for $x \neq a$, then take the limit at the end.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
Three techniques for resolving $\frac{0}{0}$:
| Situation | Move | Example form |
|---|---|---|
| Polynomial or rational | Factor; cancel common factor | $\dfrac{x^2 - 9}{x - 3}$: factor numerator, cancel $(x-3)$ |
| Square root in numerator or denominator | Multiply by conjugate | $\dfrac{\sqrt{x+4} - 2}{x}$: multiply by $\dfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}$ |
| Compound fraction | Combine over common denominator first | $\dfrac{\frac{1}{x+h} - \frac{1}{x}}{h}$: get LCD, simplify |
After the algebraic step, the shared factor is gone, and direct substitution finishes the job.
Key Concepts
1. Factoring and Canceling
If $f(x) = \dfrac{p(x)}{q(x)}$ and both $p(a) = 0$ and $q(a) = 0$, then $(x - a)$ is a factor of both numerator and denominator. Factor, cancel, and substitute.
The cancellation is valid for $x \neq a$. The original function may be undefined at $x = a$ (because the original denominator is zero there), but the limit asks what happens as $x$ approaches $a$, not at $a$. The canceled expression agrees with the original everywhere except at $x = a$, so their limits are identical.
Example 1. Evaluate $\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.
Direct substitution: $\frac{0}{0}$. Factor: \[ \frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x + 3 \quad (x \neq 3). \] Now substitute: $\lim_{x \to 3}(x + 3) = 6$.
Example 2. Evaluate $\displaystyle\lim_{x \to -2} \frac{x^2 + x - 2}{x^2 + 5x + 6}$.
Direct substitution: numerator $= 4 - 2 - 2 = 0$, denominator $= 4 - 10 + 6 = 0$. Factor both: \[ \frac{(x+2)(x-1)}{(x+2)(x+3)} = \frac{x - 1}{x + 3} \quad (x \neq -2). \] Substitute: $\dfrac{-2 - 1}{-2 + 3} = \dfrac{-3}{1} = -3$.
Example 3. Evaluate $\displaystyle\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$.
Recall the difference of cubes: $x^3 - 8 = (x-2)(x^2 + 2x + 4)$. \[ \frac{(x-2)(x^2 + 2x + 4)}{x-2} = x^2 + 2x + 4 \quad (x \neq 2). \] Substitute: $4 + 4 + 4 = 12$.
2. Rationalizing by Conjugate Multiplication
When the indeterminate form involves a square root, multiplying numerator and denominator by the conjugate turns the radical expression into a difference of squares, which factors and cancels.
The conjugate of $\sqrt{A} - B$ is $\sqrt{A} + B$, and their product is $A - B^2$, a polynomial.
Example 4. Evaluate $\displaystyle\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$.
Direct substitution: $\frac{0}{0}$. Multiply by conjugate: \[ \frac{\sqrt{x+9} - 3}{x} \cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3} = \frac{(x+9) - 9}{x(\sqrt{x+9}+3)} = \frac{x}{x(\sqrt{x+9}+3)} = \frac{1}{\sqrt{x+9}+3}. \] Now substitute $x = 0$: $\dfrac{1}{\sqrt{9}+3} = \dfrac{1}{6}$.
Example 5. Evaluate $\displaystyle\lim_{h \to 0} \frac{\sqrt{4 + h} - 2}{h}$.
This is the definition of the derivative of $\sqrt{x}$ at $x = 4$. Multiply by conjugate: \[ \frac{\sqrt{4+h}-2}{h} \cdot \frac{\sqrt{4+h}+2}{\sqrt{4+h}+2} = \frac{(4+h)-4}{h(\sqrt{4+h}+2)} = \frac{h}{h(\sqrt{4+h}+2)} = \frac{1}{\sqrt{4+h}+2}. \] Substitute $h = 0$: $\dfrac{1}{2+2} = \dfrac{1}{4}$.
3. Combining Fractions over a Common Denominator
Some limits have a compound fraction as the expression. Before you can cancel, you need to rewrite the compound fraction as a single fraction.
Example 6. Evaluate $\displaystyle\lim_{x \to 0} \frac{\frac{1}{x+3} - \frac{1}{3}}{x}$.
Direct substitution: numerator $= \frac{1}{3} - \frac{1}{3} = 0$, denominator $= 0$. Combine the numerator over a common denominator of $3(x+3)$: \[ \frac{1}{x+3} - \frac{1}{3} = \frac{3 - (x+3)}{3(x+3)} = \frac{-x}{3(x+3)}. \] Now the full expression is: \[ \frac{\dfrac{-x}{3(x+3)}}{x} = \frac{-x}{3(x+3) \cdot x} = \frac{-1}{3(x+3)}. \] Substitute $x = 0$: $\dfrac{-1}{3 \cdot 3} = -\dfrac{1}{9}$.
This limit is the definition of the derivative of $f(x) = 1/x$ at $x = 3$; you would get the same answer from the power rule ($f'(x) = -1/x^2$, so $f'(3) = -1/9$). Seeing the connection reinforces that derivative computations are limit computations.
4. Which Technique to Choose
| Symptom after substitution gives $\frac{0}{0}$ | Technique |
|---|---|
| Expression is a ratio of polynomials | Factor both; cancel common root |
| Expression has a square root subtracted from or added to a polynomial | Multiply numerator and denominator by conjugate |
| Expression has fractions inside a fraction | Combine numerator fractions over LCD; cancel |
| None of the above works | Try the squeeze theorem or revisit factoring |
In practice: factor first (it is usually fastest). If the numerator or denominator has a square root, conjugate multiplication. If there are fractions inside, LCD first.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Canceling before factoring | “Cancel an $x$” in $\dfrac{x^2 - 4}{x}$ -- wrong if $x$ is not a common factor of numerator | Factor fully: $\dfrac{x(x) - 4}{x}$ -- the $x$’s do not cancel because $-4$ prevents factoring $x$ |
| Forgetting the $\pm$ when taking the square root | Writing $\sqrt{(x-2)^2} = x - 2$ instead of $|x-2|$ | Take absolute value, then use the sign based on the limit direction |
| Canceling on only one side | $\dfrac{x(x-3)}{x+3}$: “cancel $x$” | There is no $x$ in the denominator to cancel; you can only cancel a factor present in both |
| Stopping after simplifying but before substituting | Writing $x + 3$ without computing the limit | The limit requires substituting $x = a$ into the simplified expression |
Leveled Practice
Level 1 -- Factor and Cancel
Problem 1. Evaluate $\displaystyle\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$.
Show answer
$x^2 - 25 = (x-5)(x+5)$. Cancel $(x-5)$: limit is $\lim_{x \to 5} (x+5) = 10$.
Problem 2. Evaluate $\displaystyle\lim_{x \to 1} \frac{x^2 - 3x + 2}{x - 1}$.
Show answer
Factor: $x^2 - 3x + 2 = (x-1)(x-2)$. Cancel: limit is $\lim_{x \to 1}(x-2) = -1$.
Problem 3. Evaluate $\displaystyle\lim_{x \to -1} \frac{x^3 + 1}{x + 1}$.
Show answer
Sum of cubes: $x^3 + 1 = (x+1)(x^2 - x + 1)$. Cancel: limit is $\lim_{x \to -1}(x^2 - x + 1) = 1 + 1 + 1 = 3$.
Level 2 -- Conjugate and Common Denominator
Problem 4. Evaluate $\displaystyle\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$.
Show answer
Multiply by conjugate $\dfrac{\sqrt{x}+2}{\sqrt{x}+2}$: \[ \frac{x - 4}{(x-4)(\sqrt{x}+2)} = \frac{1}{\sqrt{x}+2}. \] Limit: $\dfrac{1}{\sqrt{4}+2} = \dfrac{1}{4}$.
Problem 5. Evaluate $\displaystyle\lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$ (treat $x$ as a fixed positive constant).
Show answer
Combine numerator: $\dfrac{1}{x+h} - \dfrac{1}{x} = \dfrac{x-(x+h)}{x(x+h)} = \dfrac{-h}{x(x+h)}$.
Divide by $h$: $\dfrac{-h}{hx(x+h)} = \dfrac{-1}{x(x+h)}$.
Limit as $h \to 0$: $\dfrac{-1}{x \cdot x} = -\dfrac{1}{x^2}$.
This is the derivative of $1/x$.
Level 3 -- Mixed and Deeper
Problem 6. Evaluate $\displaystyle\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}$.
Show answer
Conjugate: multiply by $\dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$: \[ \frac{(1+x)-1}{x(\sqrt{1+x}+1)} = \frac{x}{x(\sqrt{1+x}+1)} = \frac{1}{\sqrt{1+x}+1}. \] Limit: $\dfrac{1}{1+1} = \dfrac{1}{2}$.
This equals the derivative of $\sqrt{x}$ at $x = 1$.
Problem 7. Evaluate $\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x^2 - 3x + 2}$.
Show answer
Factor both: numerator $= (x-2)(x+2)$, denominator $= (x-2)(x-1)$. Cancel: \[ \lim_{x \to 2} \frac{x+2}{x-1} = \frac{4}{1} = 4. \]
Problem 8. Evaluate $\displaystyle\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}$.
Show answer
Factor numerator as a difference of squares in $\sqrt{x}$: $x - 9 = (\sqrt{x}-3)(\sqrt{x}+3)$. Cancel $(\sqrt{x}-3)$: \[ \lim_{x \to 9} (\sqrt{x}+3) = 3 + 3 = 6. \]
Common Misconceptions
canceling a factor removes a restriction on the domain. When students simplify $\frac{(x-2)(x+3)}{x-2}$ to $x+3$, they sometimes conclude the two expressions are identical. They are equal for all $x \neq 2$, but the original expression is undefined at $x = 2$ while $x + 3$ gives 5 there. The cancellation is valid for computing $\lim_{x \to 2}$ because the limit ignores the single point $x = 2$, but it does not change the domain of the original function.
every $0/0$ form means the limit does not exist. The form $0/0$ is called indeterminate because it does not determine the limit. It simply means direct substitution failed. After algebraic manipulation, the limit may turn out to be any finite number, $+\infty$, $-\infty$, or genuinely nonexistent. The label $0/0$ is a signal to try factoring, rationalizing, or another algebraic technique, not a conclusion about the limit itself.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Every algebraic technique here is the same act: rewrite the expression so the shared zero is visible as a factor, then cancel it.
The limit does not care what happens at $x = a$; it cares about $x$ near $a$. So you are free to rewrite the expression in a different algebraic form, as long as the new form agrees with the old form for $x \neq a$. Canceling $(x - a)$ from numerator and denominator is safe for $x \neq a$, which is all the limit needs.
After the rewrite, what remains is a function that is continuous at $a$, so direct substitution finishes the job.
Connections
Within Calculus I (MATH161)
- Direct substitution: The algebraic techniques here are what you use when direct substitution gives $\frac{0}{0}$. Direct substitution is always the first attempt.
- Derivative definition: $\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$ always gives $\frac{0}{0}$ at $h = 0$. Every derivative by first principles is an algebraic-technique limit.
- Squeeze theorem: When factoring and conjugates both fail, the squeeze theorem may apply.
Toward Later Courses
- L’Hopital’s Rule (MATH162): An alternative to algebra for $\frac{0}{0}$ and $\frac{\infty}{\infty}$ that uses derivatives. The algebraic approach taught here gives exact insight into why the limit exists.
Back to Calculus I Skills | Previous: Direct Substitution | Next: One-Sided Limit Calculation