Test questions for Pre-Calculus: Rationalizing limits algebraically

📚 Quick Study Guide

🔍 When evaluating limits, if direct substitution results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, algebraic manipulation is required. 💡 Rationalization is a technique used to eliminate radicals (usually square roots) from the numerator or denominator of a fraction. 📝 Multiply the numerator and denominator by the conjugate of the expression containing the radical. ➕ The conjugate of $a + b$ is $a - b$, and vice-versa. Remember that $(a+b)(a-b) = a^2 - b^2$. ➗ After rationalizing, simplify the expression by canceling out common factors. 📈 If direct substitution still results in an indeterminate form, repeat the process or consider other algebraic techniques. 📌 Always check your answer by plugging in values close to the limit point to verify the result.

Practice Quiz

1. What is the first step in rationalizing the limit $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$?
1. Multiply the numerator and denominator by $\sqrt{x} - 2$.
2. Multiply the numerator and denominator by $\sqrt{x} + 2$.
3. Directly substitute $x = 4$.
4. Factor the denominator.
2. Evaluate the limit $\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}$.
1. 0
2. 3
3. 6
4. Does not exist.
3. What is the conjugate of $\sqrt{x+5} + \sqrt{5}$?
1. $\sqrt{x+5} + \sqrt{5}$
2. $\sqrt{x+5} - \sqrt{5}$
3. $-\sqrt{x+5} - \sqrt{5}$
4. $\sqrt{x+5} + 5$
4. Evaluate the limit $\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x}$.
1. 0
2. $\frac{1}{4}$
3. $\frac{1}{2}$
4. Does not exist.
5. Which expression is obtained after rationalizing the numerator of $\frac{\sqrt{x+1} - 1}{x}$?
1. $\frac{x}{(x(\sqrt{x+1} + 1))}$
2. $\frac{x+2}{x(\sqrt{x+1} + 1)}$
3. $\frac{x}{(\sqrt{x+1} + 1)}$
4. $\frac{1}{\sqrt{x+1} + 1}$
6. Find the limit $\lim_{x \to 2} \frac{\sqrt{6-x} - 2}{\sqrt{2-x}}$.
1. 0
2. 1
3. -1
4. Does not exist.
7. Evaluate $\lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$.
1. $\frac{1}{2\sqrt{x}}$
2. $2\sqrt{x}$
3. $\frac{1}{\sqrt{x}}$
4. $\sqrt{x}$