`lim_(x to pi) (sqrt(2 + cos x)-1)/(pi-x)^(2)`

Options

A. `2/pi`
B. `pi`
C. `1/4`
D. `1/2`

Answer

To solve the limit \( \lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{(\pi - x)^2} \), we can follow these steps:

### Step 1: Direct Substitution
First, we substitute \( x = \pi \) into the expression:

\[ \sqrt{2 + \cos(\pi)} - 1 = \sqrt{2 - 1} - 1 = \sqrt{1} - 1 = 0 \]

And the denominator:

\[ (\pi - \pi)^2 = 0^2 = 0 \]

Thus, we have the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator.

#### Derivative of the Numerator
Let \( f(x) = \sqrt{2 + \cos x} - 1 \). The derivative \( f'(x) \) is:

\[ f'(x) = \frac{1}{2\sqrt{2 + \cos x}} \cdot (-\sin x) \]

So,

\[ f'(x) = -\frac{\sin x}{2\sqrt{2 + \cos x}} \]

#### Derivative of the Denominator
Let \( g(x) = (\pi - x)^2 \). The derivative \( g'(x) \) is:

\[ g'(x) = 2(\pi - x)(-1) = -2(\pi - x) \]

### Step 3: Rewrite the Limit
Now we can rewrite the limit using L'Hôpital's Rule:

\[ \lim_{x \to \pi} \frac{f'(x)}{g'(x)} = \lim_{x \to \pi} \frac{-\frac{\sin x}{2\sqrt{2 + \cos x}}}{-2(\pi - x)} \]

This simplifies to:

\[ \lim_{x \to \pi} \frac{\sin x}{4(\pi - x)\sqrt{2 + \cos x}} \]

### Step 4: Substitute Again
Now we substitute \( x = \pi \):

\[ \sin(\pi) = 0 \]

And the denominator:

\[ 4(\pi - \pi)\sqrt{2 + \cos(\pi)} = 4(0)\sqrt{2 - 1} = 0 \]

We still have the form \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 5: Second Application of L'Hôpital's Rule
We need to differentiate the numerator and denominator again.

#### Second Derivative of the Numerator
The numerator is \( \sin x \), so its derivative is:

\[ \cos x \]

#### Second Derivative of the Denominator
The denominator is \( 4(\pi - x)\sqrt{2 + \cos x} \). Using the product rule:

\[ g'(x) = 4\left(-\sqrt{2 + \cos x} + (\pi - x)\cdot \frac{-\sin x}{2\sqrt{2 + \cos x}}\right) \]

### Step 6: Rewrite the Limit Again
Now we can rewrite the limit:

\[ \lim_{x \to \pi} \frac{\cos x}{4\left(-\sqrt{2 + \cos x} + (\pi - x)\cdot \frac{-\sin x}{2\sqrt{2 + \cos x}}\right)} \]

### Step 7: Final Substitution
Substituting \( x = \pi \):

\[ \cos(\pi) = -1 \quad \text{and} \quad \sqrt{2 + \cos(\pi)} = \sqrt{1} = 1 \]

Thus, we have:

\[ \lim_{x \to \pi} \frac{-1}{4\left(-1 + 0\right)} = \frac{-1}{-4} = \frac{1}{4} \]

### Final Answer
Therefore, the limit is:

\[ \lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{(\pi - x)^2} = \frac{1}{4} \]

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