1. The Rationalization Method
Rationalization is a technique used to evaluate a limit when direct substitution results in an indeterminate form, usually \( \frac{0}{0} \), and the function contains a square root. The goal is to eliminate the root from the numerator or denominator by multiplying by the **conjugate**.
Conjugates:
- The conjugate of \( \sqrt{a} - b \) is \( \sqrt{a} + b \).
- The conjugate of \( \sqrt{a} + b \) is \( \sqrt{a} - b \).
Multiplying a binomial by its conjugate uses the difference of squares formula: \( (a - b)(a + b) = a^2 - b^2 \). This will help you get rid of the square root!
2. Example Problem
Example: Find the limit by rationalization.
\[ \lim_{x \to 0} \frac{\sqrt{x+9} - 3}{x} \]Solution:
First, direct substitution gives \( \frac{\sqrt{0+9} - 3}{0} = \frac{3-3}{0} = \frac{0}{0} \), which is an indeterminate form.
Next, we multiply the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{x+9} + 3 \).
\[ \lim_{x \to 0} \left( \frac{\sqrt{x+9} - 3}{x} \cdot \frac{\sqrt{x+9} + 3}{\sqrt{x+9} + 3} \right) \]Simplify the numerator using the difference of squares formula \( (a-b)(a+b) = a^2 - b^2 \).
\[ = \lim_{x \to 0} \frac{(\sqrt{x+9})^2 - 3^2}{x(\sqrt{x+9} + 3)} = \lim_{x \to 0} \frac{(x+9) - 9}{x(\sqrt{x+9} + 3)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+9} + 3)} \]Now, cancel the common factor \( x \).
\[ = \lim_{x \to 0} \frac{1}{\sqrt{x+9} + 3} \]Finally, substitute \( x=0 \) to evaluate the limit.
\[ = \frac{1}{\sqrt{0+9} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6} \]3. Check Yourself
Practice these problems on your own and then check the solutions below.
Question 1: Find the limit.
\[ \lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} \]\[ = \lim_{x \to 0} \frac{(x+4) - 4}{x(\sqrt{x+4} + 2)} \]
\[ = \lim_{x \to 0} \frac{x}{x(\sqrt{x+4} + 2)} \]
\[ = \lim_{x \to 0} \frac{1}{\sqrt{x+4} + 2} \]
\[ = \frac{1}{\sqrt{4} + 2} = \frac{1}{2+2} = \frac{1}{4} \]
Question 2: Find the limit.
\[ \lim_{x \to 16} \frac{\sqrt{x} - 4}{x - 16} \]\[ = \lim_{x \to 16} \frac{x - 16}{(x - 16)(\sqrt{x} + 4)} \]
\[ = \lim_{x \to 16} \frac{1}{\sqrt{x} + 4} \]
\[ = \frac{1}{\sqrt{16} + 4} = \frac{1}{4+4} = \frac{1}{8} \]
Question 3: Find the limit.
\[ \lim_{x \to 4} \frac{x-4}{\sqrt{x}-2} \]This is a common variation where the conjugate is in the denominator.
\[ \lim_{x \to 4} \frac{x-4}{\sqrt{x}-2} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2} \]\[ = \lim_{x \to 4} \frac{(x-4)(\sqrt{x}+2)}{x-4} \]
\[ = \lim_{x \to 4} (\sqrt{x}+2) \]
\[ = \sqrt{4} + 2 = 2+2 = 4 \]