The problem presented in the image involves finding the limit:

$\lim_{{x \to 2}} \frac{x^2 - 4}{\sqrt{3x - 2} - \sqrt{x + 2}}$

Step-by-Step Solution:

1. Simplify the numerator: The expression in the numerator is $x^2 - 4$, which can be factored: $x^2 - 4 = (x - 2)(x + 2)$

   Now the limit becomes: $\lim_{{x \to 2}} \frac{(x - 2)(x + 2)}{\sqrt{3x - 2} - \sqrt{x + 2}}$

2. Check direct substitution: If we directly substitute $x = 2$, we get: $\frac{(2 - 2)(2 + 2)}{\sqrt{3(2) - 2} - \sqrt{2 + 2}} = \frac{0 \cdot 4}{\sqrt{6 - 2} - \sqrt{4}} = \frac{0}{\sqrt{4} - \sqrt{4}} = \frac{0}{0}$ This is an indeterminate form, so we need to apply an algebraic technique to simplify further.

3. Multiply by the conjugate: To deal with the indeterminate form, multiply both the numerator and denominator by the conjugate of the denominator: $\frac{\sqrt{3x - 2} - \sqrt{x + 2}}{\sqrt{3x - 2} - \sqrt{x + 2}} \cdot \frac{\sqrt{3x - 2} + \sqrt{x + 2}}{\sqrt{3x - 2} + \sqrt{x + 2}}$

   This simplifies the denominator using the difference of squares: $\lim_{{x \to 2}} \frac{(x - 2)(x + 2) \cdot (\sqrt{3x - 2} + \sqrt{x + 2})}{(3x - 2) - (x + 2)}$

4. Simplify the denominator: Simplify the expression in the denominator: $(3x - 2) - (x + 2) = 3x - 2 - x - 2 = 2x - 4 = 2(x - 2)$ Now the limit is: $\lim_{{x \to 2}} \frac{(x - 2)(x + 2) \cdot (\sqrt{3x - 2} + \sqrt{x + 2})}{2(x - 2)}$

5. Cancel the common factor: Cancel out the common factor of $x - 2$ in the numerator and denominator: $\lim_{{x \to 2}} \frac{(x + 2) \cdot (\sqrt{3x - 2} + \sqrt{x + 2})}{2}$

6. Substitute $x = 2$: Now substitute $x = 2$ into the simplified expression: $\frac{(2 + 2) \cdot (\sqrt{3(2) - 2} + \sqrt{2 + 2})}{2} = \frac{4 \cdot (\sqrt{6 - 2} + \sqrt{4})}{2} = \frac{4 \cdot (\sqrt{4} + 2)}{2} = \frac{4 \cdot (2 + 2)}{2} = \frac{4 \cdot 4}{2} = \frac{16}{2} = 8$

Final Answer:

$\lim_{{x \to 2}} \frac{x^2 - 4}{\sqrt{3x - 2} - \sqrt{x + 2}} = 8$

Would you like any more details on this solution or have any other questions?

Related Questions:

1. How does multiplying by the conjugate help simplify limits involving square roots?
2. What are other techniques to solve limits involving indeterminate forms?
3. How can L'Hopital's Rule be applied to solve limits like this one?
4. Can this problem be solved graphically, and how would the result compare?
5. What are common cases where limits result in an indeterminate form?

Tip:

When simplifying limits involving square roots, multiplying by the conjugate is often a very effective technique to eliminate the radicals and simplify the expression.