` lim_(xto pi/4) (cot^3x-tanx)/(cos(x+pi/4))` is

Options

A. 4
B. `8sqrt2`
C. 8
D. `4sqrt2`

Answer

To solve the limit problem \( \lim_{x \to \frac{\pi}{4}} \frac{\cot^3 x - \tan x}{\cos(x + \frac{\pi}{4})} \), we will follow these steps:

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

\[ \cot\left(\frac{\pi}{4}\right) = 1 \quad \text{and} \quad \tan\left(\frac{\pi}{4}\right) = 1 \]
Thus,
\[ \cot^3\left(\frac{\pi}{4}\right) - \tan\left(\frac{\pi}{4}\right) = 1^3 - 1 = 0 \]

Next, we calculate the denominator:
\[ \cos\left(\frac{\pi}{4} + \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]

So we have:
\[ \frac{0}{0} \]
This indicates that we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
We differentiate the numerator and the denominator separately.

**Numerator:**
\[ \frac{d}{dx}(\cot^3 x - \tan x) = 3\cot^2 x \cdot (-\csc^2 x) - \sec^2 x \]
This simplifies to:
\[ -3\cot^2 x \csc^2 x - \sec^2 x \]

**Denominator:**
\[ \frac{d}{dx}(\cos(x + \frac{\pi}{4})) = -\sin(x + \frac{\pi}{4}) \]

### Step 3: Rewrite the Limit
Now we rewrite the limit using the derivatives:
\[ \lim_{x \to \frac{\pi}{4}} \frac{-3\cot^2 x \csc^2 x - \sec^2 x}{-\sin(x + \frac{\pi}{4})} \]

### Step 4: Substitute Again
Now we substitute \( x = \frac{\pi}{4} \) again:
\[ \cot\left(\frac{\pi}{4}\right) = 1, \quad \csc\left(\frac{\pi}{4}\right) = \sqrt{2}, \quad \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \]
Thus,
\[ -3(1^2)(\sqrt{2})^2 - (\sqrt{2})^2 = -3 \cdot 2 - 2 = -6 - 2 = -8 \]

For the denominator:
\[ -\sin\left(\frac{\pi}{4} + \frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{2}\right) = -1 \]

### Step 5: Final Calculation
Now we have:
\[ \lim_{x \to \frac{\pi}{4}} \frac{-8}{-1} = 8 \]

Thus, the final answer is:
\[ \boxed{8} \]