How to Evaluate a Limit by Rationalizing

Problem

\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}

Answer

\frac{1}{6}

A square root difference cannot be factored in the usual way. Multiplying by the conjugate turns it into a difference of squares, which cancels the x.

Step-by-step solution

Try direct substitution.
$\frac{\sqrt{9} - 3}{0} = \frac{0}{0}$
Why: The 0/0 form means the expression needs rewriting before you substitute.
Multiply the top and bottom by the conjugate.
$\frac{\sqrt{x + 9} - 3}{x} \cdot \frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3}$
Why: The conjugate turns the difference of roots into a difference of squares.
Simplify the numerator.
$(x + 9) - 9 = x$
Why: The square root and its conjugate multiply to the radicand minus 9.
Cancel the common factor x.
$\frac{x}{x\left(\sqrt{x + 9} + 3\right)} = \frac{1}{\sqrt{x + 9} + 3}$
Why: The factor x is common to the top and bottom for every x except 0.
Substitute x = 0 into the simplified expression.
$\frac{1}{\sqrt{9} + 3} = \frac{1}{6}$
Why: The simplified expression is defined at 0.

Common mistakes

Multiplying only the numerator by the conjugate. The denominator needs it too.
Forgetting to change the sign in the conjugate. The conjugate pairs a minus with a plus.
Substituting 0 while the x is still in the denominator.

Practice problems

Use the same method. Work each problem on paper, then open the answer to check.

$\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$

Show answer

$\frac{1}{4}$
$\lim_{x \to 0} \frac{\sqrt{x + 16} - 4}{x}$

Show answer

$\frac{1}{8}$
$\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x}$

Show answer

$\frac{1}{2}$
$\lim_{x \to 0} \frac{\sqrt{x + 25} - 5}{x}$

Show answer

$\frac{1}{10}$
$\lim_{x \to 0} \frac{\sqrt{x + 36} - 6}{x}$

Show answer

$\frac{1}{12}$