Evaluate `underset(x to pi/4)lim (1-sin 2x)/(1+cos 4x)`

Options

A. `1/4`
B. `1/2`
C. `1/8`
D. `1/(16)`

Answer

To evaluate the limit

\[ \lim_{x \to \frac{\pi}{4}} \frac{1 - \sin(2x)}{1 + \cos(4x)}, \]

we will follow these steps:

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

First, we substitute \( x = \frac{\pi}{4} \) into the expression:

\[ 1 - \sin(2 \cdot \frac{\pi}{4}) = 1 - \sin(\frac{\pi}{2}) = 1 - 1 = 0, \]

and

\[ 1 + \cos(4 \cdot \frac{\pi}{4}) = 1 + \cos(\pi) = 1 - 1 = 0. \]

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

### Step 2: Apply L'Hôpital's Rule

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

\[ \lim_{x \to \frac{\pi}{4}} \frac{1 - \sin(2x)}{1 + \cos(4x)} = \lim_{x \to \frac{\pi}{4}} \frac{\frac{d}{dx}(1 - \sin(2x))}{\frac{d}{dx}(1 + \cos(4x))}. \]

### Step 3: Differentiate the Numerator and Denominator

Now we differentiate the numerator and denominator:

- The derivative of the numerator \( 1 - \sin(2x) \) is:

\[ -\cos(2x) \cdot 2 = -2\cos(2x). \]

- The derivative of the denominator \( 1 + \cos(4x) \) is:

\[ -\sin(4x) \cdot 4 = -4\sin(4x). \]

### Step 4: Rewrite the Limit

Now, we rewrite the limit:

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

### Step 5: Substitute \( x = \frac{\pi}{4} \) Again

Now we substitute \( x = \frac{\pi}{4} \) again:

- For the numerator:

\[ \cos(2 \cdot \frac{\pi}{4}) = \cos(\frac{\pi}{2}) = 0, \]

- For the denominator:

\[ \sin(4 \cdot \frac{\pi}{4}) = \sin(\pi) = 0. \]

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

### Step 6: Differentiate Again

We differentiate again:

- The derivative of the numerator \( \cos(2x) \) is:

\[ -\sin(2x) \cdot 2 = -2\sin(2x). \]

- The derivative of the denominator \( 2\sin(4x) \) is:

\[ 2 \cdot 4\cos(4x) = 8\cos(4x). \]

### Step 7: Rewrite the Limit Again

Now we rewrite the limit:

\[ \lim_{x \to \frac{\pi}{4}} \frac{-2\sin(2x)}{8\cos(4x)} = \lim_{x \to \frac{\pi}{4}} \frac{-\sin(2x)}{4\cos(4x)}. \]

### Step 8: Substitute \( x = \frac{\pi}{4} \) Once More

Now we substitute \( x = \frac{\pi}{4} \):

- For the numerator:

\[ -\sin(2 \cdot \frac{\pi}{4}) = -\sin(\frac{\pi}{2}) = -1, \]

- For the denominator:

\[ 4\cos(4 \cdot \frac{\pi}{4}) = 4\cos(\pi) = 4 \cdot (-1) = -4. \]

### Step 9: Calculate the Limit

Now we calculate the limit:

\[ \frac{-1}{-4} = \frac{1}{4}. \]

Thus, the final answer is:

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