Math Problem Statement

Find the limit of the following: lim_{x→2} (2 - √(x + 2)) / (x^2 - 4). Use decimal numbers only. Keep up to 4 decimal points.

Solution

To solve this limit, we first observe the given expression:

limx22x+2x24.\lim_{x \to 2} \frac{2 - \sqrt{x + 2}}{x^2 - 4}.

Step-by-step solution:

  1. Substitute x=2x = 2: If we directly substitute x=2x = 2: 22+2224=2444=220=00,\frac{2 - \sqrt{2 + 2}}{2^2 - 4} = \frac{2 - \sqrt{4}}{4 - 4} = \frac{2 - 2}{0} = \frac{0}{0}, which is an indeterminate form.

  2. Simplify the numerator: Notice that the numerator 2x+22 - \sqrt{x + 2} can be rewritten using a rationalizing technique: 2x+2=(2x+2)(2+x+2)2+x+2=4(x+2)2+x+2.2 - \sqrt{x + 2} = \frac{(2 - \sqrt{x + 2})(2 + \sqrt{x + 2})}{2 + \sqrt{x + 2}} = \frac{4 - (x + 2)}{2 + \sqrt{x + 2}}. Simplify 4(x+2)4 - (x + 2): 4(x+2)=2x.4 - (x + 2) = 2 - x. Therefore, the numerator becomes: 2x+2=2x2+x+2.2 - \sqrt{x + 2} = \frac{2 - x}{2 + \sqrt{x + 2}}.

  3. Substitute back into the original limit: Replace the numerator in the original fraction: limx22x+2x24=limx22x2+x+2x24.\lim_{x \to 2} \frac{2 - \sqrt{x + 2}}{x^2 - 4} = \lim_{x \to 2} \frac{\frac{2 - x}{2 + \sqrt{x + 2}}}{x^2 - 4}.

  4. Simplify the overall fraction: Rewrite the division: 2x2+x+2x24=2x(2+x+2)(x24).\frac{\frac{2 - x}{2 + \sqrt{x + 2}}}{x^2 - 4} = \frac{2 - x}{(2 + \sqrt{x + 2})(x^2 - 4)}. Notice that x24=(x2)(x+2)x^2 - 4 = (x - 2)(x + 2), so: 2x(2+x+2)(x2)(x+2).\frac{2 - x}{(2 + \sqrt{x + 2})(x - 2)(x + 2)}.

  5. Simplify further: Since 2x=(x2)2 - x = -(x - 2), the expression simplifies to: (x2)(2+x+2)(x2)(x+2)=1(2+x+2)(x+2).\frac{-(x - 2)}{(2 + \sqrt{x + 2})(x - 2)(x + 2)} = \frac{-1}{(2 + \sqrt{x + 2})(x + 2)}.

  6. Take the limit as x2x \to 2: Substitute x=2x = 2 into the simplified expression: 1(2+2+2)(2+2)=1(2+2)(4)=116.\frac{-1}{(2 + \sqrt{2 + 2})(2 + 2)} = \frac{-1}{(2 + 2)(4)} = \frac{-1}{16}.

Final Answer:

0.0625\boxed{-0.0625}

Would you like further clarification or a graphical representation? Let me know!

Follow-up questions:

  1. What happens if we solve this problem using numerical approaches?
  2. Can this limit be solved without rationalization?
  3. How does substituting values close to x=2x = 2 help in verifying this result?
  4. Can we rewrite the limit using L'Hopital's rule?
  5. What are other common indeterminate forms, and how are they handled?

Tip:

Always check for indeterminate forms (00\frac{0}{0}, \infty - \infty, etc.) when solving limits. Rationalizing or factoring are key techniques!

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Math Problem Analysis

Mathematical Concepts

Limits
Algebraic Simplification
Indeterminate Forms
Rationalization

Formulas

Rationalizing the numerator
Limit definition

Theorems

Properties of limits
Techniques to handle indeterminate forms (e.g., rationalization)

Suitable Grade Level

Grades 11-12 (Calculus)