If `alpha` is a positive root of `p(x)= x^2-x-2` Find `lim_(x rarr alpha^+) (sqrt(1-cos(p(x))))/(x-2)`

Answer

To solve the limit problem given in the question, we will follow these steps:

### Step 1: Identify the positive root \( \alpha \)
The polynomial given is \( p(x) = x^2 - x - 2 \). To find the roots, we can factor or use the quadratic formula.

The roots can be calculated as follows:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -1, c = -2 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot (-2) = 1 + 8 = 9 \]
Thus, the roots are:
\[ x = \frac{1 \pm 3}{2} = \{2, -1\} \]
The positive root \( \alpha \) is \( 2 \).

### Step 2: Rewrite the limit
We need to evaluate:
\[ \lim_{x \to \alpha^+} \frac{\sqrt{1 - \cos(p(x))}}{x - 2} \]
Substituting \( \alpha = 2 \):
\[ \lim_{x \to 2^+} \frac{\sqrt{1 - \cos(p(x))}}{x - 2} \]

### Step 3: Evaluate \( p(x) \) at \( x = 2 \)
Calculating \( p(2) \):
\[ p(2) = 2^2 - 2 - 2 = 0 \]
Thus, as \( x \to 2^+ \), \( p(x) \to 0 \).

### Step 4: Apply the limit
We have a \( \frac{0}{0} \) form, so we can apply L'Hôpital's Rule. First, we need to express \( \sqrt{1 - \cos(p(x))} \) in terms of \( p(x) \):
Using the small angle approximation:
\[ 1 - \cos(\theta) \approx \frac{\theta^2}{2} \text{ as } \theta \to 0 \]
Thus,
\[ 1 - \cos(p(x)) \approx \frac{(p(x))^2}{2} \]
Therefore,
\[ \sqrt{1 - \cos(p(x))} \approx \sqrt{\frac{(p(x))^2}{2}} = \frac{|p(x)|}{\sqrt{2}} \]

### Step 5: Substitute back into the limit
Now substituting this back into the limit:
\[ \lim_{x \to 2^+} \frac{\frac{|p(x)|}{\sqrt{2}}}{x - 2} \]
Since \( p(x) = (x - 2)(x + 1) \), we can write:
\[ |p(x)| = |(x - 2)(x + 1)| \]
As \( x \to 2^+ \), \( x - 2 \) is positive, so:
\[ |p(x)| = (x - 2)(x + 1) \]
Thus, the limit becomes:
\[ \lim_{x \to 2^+} \frac{(x - 2)(x + 1)}{\sqrt{2}(x - 2)} = \lim_{x \to 2^+} \frac{x + 1}{\sqrt{2}} \]

### Step 6: Evaluate the limit
Now substituting \( x = 2 \):
\[ \frac{2 + 1}{\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \]

### Final Answer
Thus, the final answer is:
\[ \frac{3\sqrt{2}}{2} \]