Do I always multiply by the conjugate when I see a square root in a limit?

No. Multiply by the conjugate only when you have an indeterminate form like \frac{0}{0} after substitution and the radical is in the numerator or denominator. If substitution gives a finite answer or the denominator is nonzero, you are done.

What if the denominator also has a square root?

You may need to rationalize both the numerator and the denominator separately, or rationalize one and check whether that is enough. Some limits require multiple rounds of rationalizing or a different technique such as L'Hopital's Rule or algebraic simplification.

Why does multiplying by the conjugate actually work?

The conjugate creates a difference of squares, a^2 - b^2 = (a-b)(a+b), which eliminates the radicals and converts the expression into one with a common factor in the numerator and denominator. Canceling this factor leaves a function that can be evaluated by direct substitution.

Does rationalizing work for cube roots or fourth roots?

Yes, but the formula changes. For cube roots, you use the sum of cubes factorization: a^3 - b^3 = (a - b)(a^2 + ab + b^2). Higher roots involve even more complex factorizations, so those cases are often easier with L'Hopital's Rule.

Limits by Rationalizing

Multiply by the conjugate of the numerator to eliminate radicals in limits, resolving 0/0 indeterminate forms so the limit can be evaluated.

\frac{\sqrt{x} - \sqrt{a}}{x - a} \cdot \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}

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What each symbol means

What Limits by Rationalizing takes

Symbol	Meaning
$x$	The variable approaching a particular value; must be treated as the active quantity that changes, distinct from the constant $a$ to avoid prematurely canceling them together.
$a$	A constant value that $x$ approaches; must not be confused with the variable $x$ itself.

When to use it

When a limit involving a square root produces the indeterminate form 0/0 after substitution.

Level

Usually taught in: Calculus I

Worked examples

1. Simple square root limit

Problem

Find

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

$\text{Check: } x = 4 \implies \frac{\sqrt{4} - 2}{4 - 4} = \frac{0}{0}$
Direct substitution produces the indeterminate form $\frac{0}{0}$ , so we must rationalize.
$\frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2}$
Multiply both numerator and denominator by the conjugate $\sqrt{x} + 2$ of the numerator.
$\frac{(\sqrt{x})^2 - 2^2}{(x - 4)(\sqrt{x} + 2)} = \frac{x - 4}{(x - 4)(\sqrt{x} + 2)}$
Apply the difference of squares formula $(a-b)(a+b) = a^2 - b^2$ with $a = \sqrt{x}$ and $b = 2$ .
$\frac{1}{\sqrt{x} + 2}$
Cancel the common factor $(x - 4)$ from numerator and denominator.
$\lim_{x \to 4} \frac{1}{\sqrt{x} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{4}$
Direct substitution now works; substitute $x = 4$ to evaluate the limit.

Answer: $\frac{1}{4}$

The conjugate removes the radical from the numerator, creating a difference of squares that produces a common factor $(x - 4)$ in both numerator and denominator, which can then be canceled, leaving a continuous function that can be evaluated.

2. Compound radical expression with factoring

Problem

Find

\lim_{x \to 1} \frac{\sqrt{3x + 1} - 2}{x - 1}

$\text{Check: } x = 1 \implies \frac{\sqrt{3(1) + 1} - 2}{1 - 1} = \frac{\sqrt{4} - 2}{0} = \frac{0}{0}$
Substituting $x = 1$ yields $\frac{0}{0}$ , an indeterminate form requiring rationalization.
$\frac{\sqrt{3x + 1} - 2}{x - 1} \cdot \frac{\sqrt{3x + 1} + 2}{\sqrt{3x + 1} + 2}$
Multiply by the conjugate $\sqrt{3x + 1} + 2$ to rationalize the numerator.
$\frac{(\sqrt{3x+1})^2 - 4}{(x-1)(\sqrt{3x+1}+2)} = \frac{3x + 1 - 4}{(x-1)(\sqrt{3x+1}+2)} = \frac{3x - 3}{(x-1)(\sqrt{3x+1}+2)}$
Apply the difference of squares $(a-b)(a+b) = a^2 - b^2$ , then simplify the numerator from $3x + 1 - 4$ to $3x - 3$ .
$\frac{3(x - 1)}{(x-1)(\sqrt{3x+1}+2)} = \frac{3}{\sqrt{3x+1}+2}$
Factor the numerator as $3x - 3 = 3(x-1)$ , then cancel $(x-1)$ from both parts.
$\lim_{x \to 1} \frac{3}{\sqrt{3x+1}+2} = \frac{3}{\sqrt{3(1)+1}+2} = \frac{3}{\sqrt{4}+2} = \frac{3}{4}$
With the common factor removed, substitute $x = 1$ to get $\frac{3}{2+2} = \frac{3}{4}$ .

Answer: $\frac{3}{4}$

Even when the radical expression is more complex, the same conjugate technique eliminates the radical and creates a factorable numerator, after which cancellation yields an evaluable form.

3. Phone plan billing limit

Problem

A data plan charges

\frac{\sqrt{x + 49} - 7}{x}

dollars for

x

gigabytes. Find the limit of the average cost per gigabyte as usage approaches zero:

\lim_{x \to 0} \frac{\sqrt{x + 49} - 7}{x}

$\text{Check: } x = 0 \implies \frac{\sqrt{0 + 49} - 7}{0} = \frac{7 - 7}{0} = \frac{0}{0}$
Substituting $x = 0$ gives $\frac{0}{0}$ , which is indeterminate.
$\frac{\sqrt{x+49}-7}{x} \cdot \frac{\sqrt{x+49}+7}{\sqrt{x+49}+7}$
Multiply numerator and denominator by the conjugate $\sqrt{x+49}+7$ .
$\frac{(\sqrt{x+49})^2 - 49}{x(\sqrt{x+49}+7)} = \frac{x + 49 - 49}{x(\sqrt{x+49}+7)} = \frac{x}{x(\sqrt{x+49}+7)}$
Apply difference of squares: $(\sqrt{x+49})^2 = x + 49$ , so the numerator simplifies to $x$ .
$\frac{1}{\sqrt{x+49}+7}$
Cancel the common factor $x$ from numerator and denominator.
$\lim_{x \to 0} \frac{1}{\sqrt{x+49}+7} = \frac{1}{\sqrt{49}+7} = \frac{1}{7+7} = \frac{1}{14}$
Substitute $x = 0$ to find the limit is $\frac{1}{14}$ dollars per gigabyte.

Answer: $\frac{1}{14}$

Rationalizing allows engineers and economists to evaluate limits in real-world contexts like pricing models, revealing the instantaneous rate of change even when the direct substitution appears undefined.

Common mistakes

Where Limits by Rationalizing usually goes wrong

The mistake	Why it is wrong	The fix
Using the wrong conjugate, such as $\sqrt{x} - 2$ instead of $\sqrt{x} + 2$ .	The conjugate must reverse the sign between the radical and constant to produce a difference of squares; using the same sign produces a sum instead, which does not eliminate the radical.	Always use $\sqrt{a} + \sqrt{b}$ as the conjugate for $\sqrt{a} - \sqrt{b}$ , ensuring the product $(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) = a - b$ removes the radicals.
Multiplying only the numerator by the conjugate and leaving the denominator unchanged.	You must multiply by $\frac{\text{conjugate}}{\text{conjugate}}$ , which equals 1; multiplying only the numerator changes the value of the expression and produces an incorrect limit.	Always write and apply the full multiplication: $\frac{\text{numerator}}{\text{denominator}} \cdot \frac{\text{conjugate}}{\text{conjugate}}$ , so both parts are multiplied by the same thing.
Canceling a factor before verifying it appears in both the numerator and denominator.	After rationalizing, the factor you plan to cancel must appear exactly in both parts; if it only appears once or in a different form, canceling produces the wrong simplified expression.	Always factor the numerator completely after rationalizing and confirm the factor appears in both numerator and denominator before canceling it.

Tips and when to use something else

Rationalizing converts a 0/0 indeterminate form into a continuous function by eliminating radicals, allowing you to cancel common factors and then substitute directly.
If rationalizing still leaves a complicated denominator with a radical, or if the algebra becomes messy, try L'Hopital's Rule instead—it may resolve the limit faster.
Always verify that the original expression produces 0/0 (not 0, ∞, or another form) before rationalizing; if direct substitution gives a finite answer, you are already done.
Double-check your algebra when expanding $(a - b)(a + b) = a^2 - b^2$ ; arithmetic mistakes in this step are the most common source of incorrect final answers.

Related formulas

Definition of a LimitDefinition of a Limit describes what value a function approaches as its input approaches a specific point—essential for all of calculus.Indeterminate FormsIndeterminate forms are expressions like 0/0 and ∞/∞ that arise when evaluating limits, signaling that direct substitution doesn't work.Limits by FactoringLimits by Factoring removes common factors from numerator and denominator to resolve 0/0 indeterminate forms and evaluate limits algebraically.L'Hopital's RuleL'Hopital's Rule efficiently solves limits that reduce to 0/0 or ∞/∞ forms by taking derivatives of the numerator and denominator separately.Limit LawsThe Sum Law shows that a limit of a sum equals the sum of the limits, provided both limits exist and are finite numbers.

Frequently asked questions

Do I always multiply by the conjugate when I see a square root in a limit?: No. Multiply by the conjugate only when you have an indeterminate form like $\frac{0}{0}$ after substitution and the radical is in the numerator or denominator. If substitution gives a finite answer or the denominator is nonzero, you are done.
What if the denominator also has a square root?: You may need to rationalize both the numerator and the denominator separately, or rationalize one and check whether that is enough. Some limits require multiple rounds of rationalizing or a different technique such as L'Hopital's Rule or algebraic simplification.
Why does multiplying by the conjugate actually work?: The conjugate creates a difference of squares, $a^2 - b^2 = (a-b)(a+b)$ , which eliminates the radicals and converts the expression into one with a common factor in the numerator and denominator. Canceling this factor leaves a function that can be evaluated by direct substitution.
Does rationalizing work for cube roots or fourth roots?: Yes, but the formula changes. For cube roots, you use the sum of cubes factorization: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ . Higher roots involve even more complex factorizations, so those cases are often easier with L'Hopital's Rule.

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Reviewed 2026-09-18