`lim_(x to pi//4)(1- tan x)/(1-sqrt(2) sin x)` equals:

Options

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

Answer

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

### Step 1: Direct Substitution
First, we substitute \( x = \frac{\pi}{4} \) into the expression.

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

Substituting these values:

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

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:

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

where \( f(x) = 1 - \tan x \) and \( g(x) = 1 - \sqrt{2} \sin x \).

### Step 3: Differentiate the Numerator and Denominator
Now we differentiate both the numerator and the denominator.

- The derivative of the numerator \( f(x) = 1 - \tan x \) is:

\[ f'(x) = -\sec^2 x \]

- The derivative of the denominator \( g(x) = 1 - \sqrt{2} \sin x \) is:

\[ g'(x) = -\sqrt{2} \cos x \]

### Step 4: Rewrite the Limit
Now we can rewrite the limit using the derivatives:

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

### Step 5: Substitute \( x = \frac{\pi}{4} \) Again
Now we substitute \( x = \frac{\pi}{4} \) into the new limit:

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

Thus,

\[ \lim_{x \to \frac{\pi}{4}} \frac{2}{\sqrt{2} \cdot \frac{\sqrt{2}}{2}} = \frac{2}{1} = 2 \]

### Final Result
Therefore, the limit is:

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