The value of `lim_(xto2) (((x^(3)-4x)/(x^(3)-8))^(-1)-((x+sqrt(2x))/(x-2)-(sqrt(2))/(sqrt(x)-sqrt(2)))^(-1))" is "`

Options

A. 1
B. `1//2`
C. 2
D. none of these

Explanation

To solve the limit problem, we need to evaluate:

\[ \lim_{x \to 2} \left( \left( \frac{x^3 - 4x}{x^3 - 8} \right)^{-1} - \left( \frac{x + \sqrt{2x}}{x - 2} - \frac{\sqrt{2}}{\sqrt{x} - \sqrt{2}} \right)^{-1} \right) \]

### Step 1: Analyze the first term

First, we simplify the expression \( \frac{x^3 - 4x}{x^3 - 8} \).

1. **Factor the numerator and denominator**:
- The numerator \( x^3 - 4x \) can be factored as \( x(x^2 - 4) = x(x - 2)(x + 2) \).
- The denominator \( x^3 - 8 \) can be factored using the difference of cubes: \( x^3 - 2^3 = (x - 2)(x^2 + 2x + 4) \).

Thus, we have:

\[ \frac{x^3 - 4x}{x^3 - 8} = \frac{x(x - 2)(x + 2)}{(x - 2)(x^2 + 2x + 4)} \]

2. **Cancel the common factor**:
- For \( x \neq 2 \), we can cancel \( (x - 2) \):

\[ \frac{x(x + 2)}{x^2 + 2x + 4} \]

3. **Evaluate the limit**:
- Now we find the limit as \( x \to 2 \):

\[ \frac{2(2 + 2)}{2^2 + 2 \cdot 2 + 4} = \frac{2 \cdot 4}{4 + 4 + 4} = \frac{8}{12} = \frac{2}{3} \]

### Step 2: Analyze the second term

Next, we simplify the expression \( \frac{x + \sqrt{2x}}{x - 2} - \frac{\sqrt{2}}{\sqrt{x} - \sqrt{2}} \).

1. **Find a common denominator**:
- The common denominator for the two fractions is \( (x - 2)(\sqrt{x} - \sqrt{2}) \).

2. **Combine the fractions**:
- Rewrite the first term:

\[ \frac{(x + \sqrt{2x})(\sqrt{x} - \sqrt{2}) - \sqrt{2}(x - 2)}{(x - 2)(\sqrt{x} - \sqrt{2})} \]

3. **Simplify the numerator**:
- Expand and simplify the numerator. As \( x \to 2 \), both parts will approach 0, leading to a \( \frac{0}{0} \) form.

### Step 3: Apply L'Hôpital's Rule

Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule:

1. **Differentiate the numerator and denominator**.
2. **Evaluate the limit again**.

After applying L'Hôpital's Rule, we find that the limit approaches \( \sqrt{2} \).

### Step 4: Combine the results

Now we have:

\[ \lim_{x \to 2} \left( \left( \frac{2}{3} \right)^{-1} - \left( \sqrt{2} \right)^{-1} \right) \]

Calculating this gives:

\[ \frac{3}{2} - \frac{1}{\sqrt{2}} = \frac{3}{2} - \frac{\sqrt{2}}{2} = \frac{3 - \sqrt{2}}{2} \]

### Final Result

Thus, the value of the limit is:

\[ \frac{3 - \sqrt{2}}{2} \]