Limits by Rationalization

Last Updated : 28 May, 2026

In calculus, a limit is a fundamental concept that describes the value that a function approaches as the input approaches a certain value. Limits help us understand the behavior of the functions at points where they might not be explicitly defined. For example, the limit of the function f(x) as x approaches a value a is denoted as the following:

\lim\limits_{x \to a} f(x)

When evaluating limits, we sometimes encounter indeterminate forms such as \frac{0}{0}​. One effective method to resolve such cases is rationalization.

Rationalization is a technique used to eliminate radicals (roots) from the numerator or denominator of an expression. This is done by multiplying and dividing by the conjugate, which helps simplify the expression and makes limit evaluation possible.

Steps to Find Limits Using Rationalization

To find limits, using rationalization, we can use the following steps:

Step 1: Identify the Indeterminate Form: If a limit results in an indeterminate form like \frac{0}{0} consider using the rationalization.

Step 2: Multiply by the Conjugate: Multiply the expression by the conjugate of the numerator or the denominator depending on where the root is present.

Step 3: Simplify the Expression: Expand and simplify the expression.

Step 4: Evaluate the Limit: After simplification substitute the limit value and calculate the limit.

Steps to Rationalize an Expression

To rationalize rational expressions, we can use the following steps:

Step 1: Identify the Expression: Recognize the limit expression that involves a radical.

Step 2: Multiply by the Conjugate: Use the conjugate of the expression to multiply the numerator and denominator.

Step 3: Simplify: Apply algebraic identities to the expression to eliminate the radical.

Step 4: Evaluate the Limit: Substitute the limit value into the simplified expression.

Solved Examples

Example 1: Evaluate \lim_{{x \to 4}} \frac{x - 4}{\sqrt{x} - 2}.

The limit initially gives an indeterminate form \frac{0}{0}. To resolve this we rationalize the numerator.

The Multiply numerator and denominator by the conjugate:

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

Simplify:

\lim_{{x \to 4}} \frac{x - 4}{ x - 4} \times \frac{\sqrt{x} + 2}{1} = \lim_{{x \to 4}} \frac{\sqrt{x} + 2}{1} = \frac{2 + 2}{1} = 4

Example 2: Find \lim_{{x \to 1}} \frac{\sqrt{x + 3} - 2}{x - 1}.

The expression gives \frac{0}{0} so we rationalize the numerator:

\lim_{{x \to 1}} \frac{\sqrt{x + 3} - 2}{x - 1} \times \frac{\sqrt{x + 3} + 2}{\sqrt{x + 3} + 2} = \lim_{{x \to 1}} \frac{(x + 3) - 4}{(x - 1)(\sqrt{x + 3} + 2)}

Simplify:

= \lim_{{x \to 1}} \frac{x - 1}{(x - 1)(\sqrt{x + 3} + 2)} = \lim_{{x \to 1}} \frac{1}{\sqrt{x + 3} + 2} = \frac{1}{\sqrt{1 + 3} + 2} = \frac{1}{4}

Example 3: Evaluate \lim_{{x \to 0}} \frac{\sqrt{1 + x} - 1}{x}.

Multiply by the conjugate:

\lim_{{x \to 0}} \frac{\sqrt{1 + x} - 1}{x} \times \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1} = \lim_{{x \to 0}} \frac{(1 + x) - 1}{x(\sqrt{1 + x} + 1)}

Simplify:

= \lim_{{x \to 0}} \frac{x}{x(\sqrt{1 + x} + 1)} = \lim_{{x \to 0}} \frac{1}{\sqrt{1 + x} + 1} = \frac{1}{2}

Example 4: Evaluate \lim_{{x \to 9}} \frac{x - 9}{\sqrt{x} - 3}.

The expression gives \frac{0}{0}. Rationalize the numerator:

\lim_{{x \to 9}} \frac{x - 9}{\sqrt{x} - 3} \times \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \lim_{{x \to 9}} \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)}

Simplify:

= \lim_{{x \to 9}} \frac{x - 9}{x - 9} \times \frac{\sqrt{x} + 3}{1} = \lim_{{x \to 9}} \frac{\sqrt{x} + 3}{1} = \frac{3 + 3}{1} = 6

Example 5: Evaluate \lim_{{x \to 0}} \frac{\sqrt{4 + x} - 2}{x}.

The Multiply by the conjugate:

\lim_{{x \to 0}} \frac{\sqrt{4 + x} - 2}{x} \times \frac{\sqrt{4 + x} + 2}{\sqrt{4 + x} + 2} = \lim_{{x \to 0}} \frac{(4 + x) - 4}{x(\sqrt{4 + x} + 2)}

Simplify:

= \lim_{{x \to 0}} \frac{x}{x(\sqrt{4 + x} + 2)} = \lim_{{x \to 0}} \frac{1}{\sqrt{4 + x} + 2} = \frac{1}{4}

Practice Problems

Problem 1: Evaluate \lim_{{x \to 16}} \frac{\sqrt{x} - 4}{x - 16}.

Problem 2: Find \lim_{{x \to 1}} \frac{\sqrt{2x + 3} - 2}{x - 1}.

Problem 3: Evaluate \lim_{{x \to 0}} \frac{\sqrt{9 + x} - 3}{x}.

Problem 4: Find \lim_{{x \to 4}} \frac{\sqrt{x + 5} - 3}{x - 4}.

Problem 5: Evaluate \lim_{{x \to 25}} \frac{\sqrt{x} - 5}{x - 25}.

Comment

Explore