`lim_(x to pi//4) (sqrt(2)cos x-1)/(cot x-1)` equals

Options

A. `1/sqrt(2)`
B. `1/2`
C. `1/(2sqrt(2))`
D. 1

Answer

To solve the limit \( \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \cos x - 1}{\cot x - 1} \), we will follow these steps:

### Step 1: Substitute \( x = \frac{\pi}{4} \)

First, we substitute \( x = \frac{\pi}{4} \) into the expression to check if it results in an indeterminate form.

\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \quad \text{and} \quad \cot\left(\frac{\pi}{4}\right) = 1 \]

Now substituting these values into the limit:

\[ \sqrt{2} \cos\left(\frac{\pi}{4}\right) - 1 = \sqrt{2} \cdot \frac{1}{\sqrt{2}} - 1 = 1 - 1 = 0 \]

\[ \cot\left(\frac{\pi}{4}\right) - 1 = 1 - 1 = 0 \]

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

### Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

if \( \lim_{x \to c} f(x) = 0 \) and \( \lim_{x \to c} g(x) = 0 \).

Let:
- \( f(x) = \sqrt{2} \cos x - 1 \)
- \( g(x) = \cot x - 1 \)

Now we need to find the derivatives \( f'(x) \) and \( g'(x) \).

### Step 3: Differentiate the Numerator and Denominator

1. **Differentiate the numerator**:
\[ f'(x) = \frac{d}{dx}(\sqrt{2} \cos x - 1) = -\sqrt{2} \sin x \]

2. **Differentiate the denominator**:
\[ g'(x) = \frac{d}{dx}(\cot x - 1) = -\csc^2 x \]

### Step 4: Rewrite the Limit

Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to \frac{\pi}{4}} \frac{f'(x)}{g'(x)} = \lim_{x \to \frac{\pi}{4}} \frac{-\sqrt{2} \sin x}{-\csc^2 x} = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \sin x}{\csc^2 x} \]

Since \( \csc x = \frac{1}{\sin x} \), we have:

\[ \csc^2 x = \frac{1}{\sin^2 x} \]

Thus, the limit becomes:

\[ \lim_{x \to \frac{\pi}{4}} \sqrt{2} \sin x \cdot \sin^2 x = \lim_{x \to \frac{\pi}{4}} \sqrt{2} \sin^3 x \]

### Step 5: Evaluate the Limit

Substituting \( x = \frac{\pi}{4} \):

\[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

So:

\[ \sqrt{2} \left(\frac{1}{\sqrt{2}}\right)^3 = \sqrt{2} \cdot \frac{1}{2\sqrt{2}} = \frac{1}{2} \]

### Final Answer

Thus, the limit is:

\[ \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \cos x - 1}{\cot x - 1} = \frac{1}{2} \]