Solved examples: Applying the conjugate method to indeterminate limits

📚 Quick Study Guide

🔍 The conjugate method is often used to evaluate limits that result in indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. 💡 The conjugate of an expression $a + b$ is $a - b$, and vice versa. Multiplying by the conjugate helps rationalize the expression. 📝 The key idea is to multiply both the numerator and the denominator by the conjugate of either the numerator or the denominator (whichever contains a square root or a term causing indeterminacy). ➗ Remember the difference of squares: $(a + b)(a - b) = a^2 - b^2$. This identity is crucial when using the conjugate method. ➕ After multiplying by the conjugate, simplify the expression and try to cancel out the terms that cause the indeterminate form. 📈 Finally, evaluate the limit by substituting the value that $x$ approaches. 💯 Practice makes perfect! Work through several examples to become comfortable with this method.

Practice Quiz

1. What is the conjugate of $\sqrt{x} + 3$?
   1. $\sqrt{x} - 3$
   2. $\sqrt{x} + 3$
   3. $-\sqrt{x} - 3$
   4. $x^2 + 9$
2. What should you multiply by to rationalize the numerator of $\frac{\sqrt{x} - 2}{x - 4}$?
   1. $\sqrt{x} - 2$
   2. $\sqrt{x} + 2$
   3. $x - 4$
   4. $x + 4$
3. Evaluate $\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$.
   1. $\frac{1}{6}$
   2. $0$
   3. $\frac{1}{3}$
   4. $\infty$
4. Evaluate $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$.
   1. $\frac{1}{4}$
   2. $0$
   3. $\frac{1}{2}$
   4. $\infty$
5. Evaluate $\lim_{x \to 1} \frac{x - 1}{\sqrt{x} - 1}$.
   1. $0$
   2. $1$
   3. $2$
   4. $\infty$
6. Evaluate $\lim_{x \to 2} \frac{\sqrt{6 - x} - 2}{\sqrt{2 - x}}$.
   1. $0$
   2. $-1$
   3. $\infty$
   4. $\frac{\sqrt{2}}{2}$
7. Find $\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x}$.
   1. $0$
   2. $1$
   3. $2$
   4. $\infty$