Math Problem Statement

Limit as x approaches 3 of (sqrt(x^2 - 5) - 2) / (2x - 6).

Solution

The limit problem from the image is:

limx3x2522x6\lim_{x \to 3} \frac{\sqrt{x^2 - 5} - 2}{2x - 6}

Step 1: Simplify the expression

Notice that when you directly substitute x=3x = 3 into the expression:

  • The denominator becomes 2(3)6=02(3) - 6 = 0
  • The numerator becomes 3252=952=42=22=0\sqrt{3^2 - 5} - 2 = \sqrt{9 - 5} - 2 = \sqrt{4} - 2 = 2 - 2 = 0

Thus, direct substitution leads to the indeterminate form 00\frac{0}{0}.

Step 2: Apply algebraic manipulation

We will rationalize the numerator to eliminate the square root. Multiply both the numerator and denominator by the conjugate of the numerator:

x2522x6×x25+2x25+2\frac{\sqrt{x^2 - 5} - 2}{2x - 6} \times \frac{\sqrt{x^2 - 5} + 2}{\sqrt{x^2 - 5} + 2}

The numerator becomes:

(x252)(x25+2)=(x25)4=x29(\sqrt{x^2 - 5} - 2)(\sqrt{x^2 - 5} + 2) = (x^2 - 5) - 4 = x^2 - 9

The denominator becomes:

(2x6)(x25+2)=(2(x3))(x25+2)(2x - 6)(\sqrt{x^2 - 5} + 2) = (2(x - 3))(\sqrt{x^2 - 5} + 2)

Thus, the expression simplifies to:

x292(x3)(x25+2)\frac{x^2 - 9}{2(x - 3)(\sqrt{x^2 - 5} + 2)}

Step 3: Factor and cancel terms

We can factor the numerator x29x^2 - 9 as:

x29=(x3)(x+3)x^2 - 9 = (x - 3)(x + 3)

Now, cancel the common factor of x3x - 3 from the numerator and denominator:

(x3)(x+3)2(x3)(x25+2)=x+32(x25+2)\frac{(x - 3)(x + 3)}{2(x - 3)(\sqrt{x^2 - 5} + 2)} = \frac{x + 3}{2(\sqrt{x^2 - 5} + 2)}

Step 4: Substitute x=3x = 3

Now that the indeterminate form has been removed, substitute x=3x = 3:

3+32(325+2)=62(95+2)=62(4+2)=62(2+2)=68=34\frac{3 + 3}{2(\sqrt{3^2 - 5} + 2)} = \frac{6}{2(\sqrt{9 - 5} + 2)} = \frac{6}{2(\sqrt{4} + 2)} = \frac{6}{2(2 + 2)} = \frac{6}{8} = \frac{3}{4}

Final Answer:

The limit is 34\frac{3}{4}, so the correct choice is B. 3/4.

Would you like further details or have any questions?

Here are some related questions for practice:

  1. How do you rationalize a numerator involving a square root?
  2. Why does direct substitution result in an indeterminate form?
  3. What is the general method to deal with indeterminate forms in limits?
  4. How does canceling common factors help simplify limits?
  5. What other techniques are useful for solving limits involving square roots?

Tip: Rationalizing the numerator is a useful technique when you encounter limits involving square roots that lead to indeterminate forms.

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Math Problem Analysis

Mathematical Concepts

Limits
Indeterminate Forms
Algebraic Manipulation
Square Root Rationalization

Formulas

Indeterminate form: 0/0
Rationalizing the numerator
Difference of squares: (a - b)(a + b) = a^2 - b^2

Theorems

L'Hopital's Rule (optional)
Limit properties for rational expressions

Suitable Grade Level

Grades 11-12 (Advanced Calculus or Precalculus)