Rationalization Method

Definition of a conjugate: For an expression of the form $a + b\sqrt{c}$, its conjugate is $a - b\sqrt{c}$.

Basic property:

• $(a + b\sqrt{c})(a - b\sqrt{c}) = a^2 - b^2c$
• This property eliminates the radical and is the core principle of rationalization.

Examples:

• The conjugate of $\sqrt{x} + 1$ is $\sqrt{x} - 1$
• The conjugate of $\sqrt{x + 1} + \sqrt{x - 1}$ is $\sqrt{x + 1} - \sqrt{x - 1}$

Memory tips:

• Square roots: change the middle sign (+ becomes −, − becomes +)
• Cube roots: use the sum and difference of cubes formulas

Idea: This is a $\frac{0}{0}$ indeterminate form containing a radical, so we can use rationalization.

Detailed steps:

1. Check the type of indeterminate form:

   • As $x \to 0$, the numerator $\sqrt{1 + x} - 1 \to 0$
   • As $x \to 0$, the denominator $x \to 0$
   • Hence it is a $\frac{0}{0}$ indeterminate form

2. Rationalize the numerator: $\frac{\sqrt{1 + x} - 1}{x} = \frac{(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1)}{x(\sqrt{1 + x} + 1)}$

3. Simplify: $\frac{(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1)}{x(\sqrt{1 + x} + 1)} = \frac{(1 + x) - 1}{x(\sqrt{1 + x} + 1)} = \frac{x}{x(\sqrt{1 + x} + 1)}$

4. Cancel $x$: $\frac{x}{x(\sqrt{1 + x} + 1)} = \frac{1}{\sqrt{1 + x} + 1}$

5. Take the limit: $\lim_{x \to 0} \frac{1}{\sqrt{1 + x} + 1} = \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{2}$

Answer: $\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x} = \frac{1}{2}$

Idea: This is a $\frac{0}{0}$ indeterminate form containing a radical, so we can use rationalization.

Detailed steps:

1. Check the type of indeterminate form:

   • When $x = 4$, the numerator $\sqrt{4} - 2 = 2 - 2 = 0$
   • When $x = 4$, the denominator $4 - 4 = 0$
   • Hence it is a $\frac{0}{0}$ indeterminate form

2. Rationalize the numerator: $\frac{\sqrt{x} - 2}{x - 4} = \frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x - 4)(\sqrt{x} + 2)}$

3. Simplify: $\frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x - 4)(\sqrt{x} + 2)} = \frac{x - 4}{(x - 4)(\sqrt{x} + 2)}$

4. Cancel the zero factor: $\frac{x - 4}{(x - 4)(\sqrt{x} + 2)} = \frac{1}{\sqrt{x} + 2}$

5. Take the limit: $\lim_{x \to 4} \frac{1}{\sqrt{x} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$

Answer: $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} = \frac{1}{4}$

Idea: This is a $\frac{0}{0}$ indeterminate form containing a radical, so we can use rationalization.

Detailed steps:

1. Check the type of indeterminate form:

   • When $x = 9$, the numerator $\sqrt{9} - 3 = 3 - 3 = 0$
   • When $x = 9$, the denominator $9 - 9 = 0$
   • Hence it is a $\frac{0}{0}$ indeterminate form

2. Rationalize the numerator: $\frac{\sqrt{x} - 3}{x - 9} = \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)}$

3. Simplify: $\frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)} = \frac{x - 9}{(x - 9)(\sqrt{x} + 3)}$

4. Cancel the zero factor: $\frac{x - 9}{(x - 9)(\sqrt{x} + 3)} = \frac{1}{\sqrt{x} + 3}$

5. Take the limit: $\lim_{x \to 9} \frac{1}{\sqrt{x} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

Answer: $\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9} = \frac{1}{6}$

Idea: This is a $\frac{0}{0}$ indeterminate form containing a radical, so we can use rationalization.

Detailed steps:

1. Check the type of indeterminate form:

   • As $x \to 0$, the numerator $\sqrt{1 + x^2} - 1 \to 0$
   • As $x \to 0$, the denominator $x^2 \to 0$
   • Hence it is a $\frac{0}{0}$ indeterminate form

2. Rationalize the numerator: $\frac{\sqrt{1 + x^2} - 1}{x^2} = \frac{(\sqrt{1 + x^2} - 1)(\sqrt{1 + x^2} + 1)}{x^2(\sqrt{1 + x^2} + 1)}$

3. Simplify: $\frac{(\sqrt{1 + x^2} - 1)(\sqrt{1 + x^2} + 1)}{x^2(\sqrt{1 + x^2} + 1)} = \frac{(1 + x^2) - 1}{x^2(\sqrt{1 + x^2} + 1)} = \frac{x^2}{x^2(\sqrt{1 + x^2} + 1)}$

4. Cancel $x^2$: $\frac{x^2}{x^2(\sqrt{1 + x^2} + 1)} = \frac{1}{\sqrt{1 + x^2} + 1}$

5. Take the limit: $\lim_{x \to 0} \frac{1}{\sqrt{1 + x^2} + 1} = \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{2}$

Answer: $\lim_{x \to 0} \frac{\sqrt{1 + x^2} - 1}{x^2} = \frac{1}{2}$