The value of `lim_(xrarr(5pi)/(4))(cot^(3)x-tanx)/(cos(x+(5pi)/(4)))` is equal to

Answer

To solve the limit

\[ \lim_{x \to \frac{5\pi}{4}} \frac{\cot^3 x - \tan x}{\cos\left(x + \frac{5\pi}{4}\right)}, \]

we will follow these steps:

### Step 1: Direct Substitution
First, we will substitute \( x = \frac{5\pi}{4} \) directly into the limit.

\[ \cot\left(\frac{5\pi}{4}\right) = \cot\left(\pi + \frac{\pi}{4}\right) = \cot\left(\frac{\pi}{4}\right) = 1, \]
\[ \tan\left(\frac{5\pi}{4}\right) = \tan\left(\pi + \frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1. \]

Thus, we have:

\[ \cot^3\left(\frac{5\pi}{4}\right) - \tan\left(\frac{5\pi}{4}\right) = 1^3 - 1 = 0. \]

Next, we evaluate the denominator:

\[ \cos\left(\frac{5\pi}{4} + \frac{5\pi}{4}\right) = \cos\left(\frac{10\pi}{4}\right) = \cos\left(\frac{5\pi}{2}\right). \]

Since \( \frac{5\pi}{2} = 2\pi + \frac{\pi}{2} \), we have:

\[ \cos\left(\frac{5\pi}{2}\right) = 0. \]

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

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

#### Differentiate the Numerator:
The numerator is \( \cot^3 x - \tan x \).

Using the chain rule and product rule:
\[ \frac{d}{dx}(\cot^3 x) = 3 \cot^2 x (-\csc^2 x) = -3 \cot^2 x \csc^2 x, \]
\[ \frac{d}{dx}(\tan x) = \sec^2 x. \]

Thus, the derivative of the numerator is:
\[ -3 \cot^2 x \csc^2 x - \sec^2 x. \]

#### Differentiate the Denominator:
The denominator is \( \cos\left(x + \frac{5\pi}{4}\right) \).

Using the chain rule:
\[ \frac{d}{dx}(\cos\left(x + \frac{5\pi}{4}\right)) = -\sin\left(x + \frac{5\pi}{4}\right). \]

### Step 3: Rewrite the Limit
Now we rewrite the limit using the derivatives:
\[ \lim_{x \to \frac{5\pi}{4}} \frac{-3 \cot^2 x \csc^2 x - \sec^2 x}{-\sin\left(x + \frac{5\pi}{4}\right)}. \]

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

For the numerator:
\[ \cot\left(\frac{5\pi}{4}\right) = 1 \Rightarrow \cot^2\left(\frac{5\pi}{4}\right) = 1, \]
\[ \csc^2\left(\frac{5\pi}{4}\right) = -2 \quad (\text{since } \sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}), \]
\[ \sec^2\left(\frac{5\pi}{4}\right) = 2 \quad (\text{since } \cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}). \]

Thus, the numerator becomes:
\[ -3(1)(-2) - 2 = 6 - 2 = 4. \]

For the denominator:
\[ \sin\left(\frac{5\pi}{4} + \frac{5\pi}{4}\right) = \sin\left(\frac{10\pi}{4}\right) = \sin\left(\frac{5\pi}{2}\right) = 1. \]

### Final Calculation
Now we can compute the limit:
\[ \lim_{x \to \frac{5\pi}{4}} \frac{4}{1} = 4. \]

Thus, the value of the limit is:

\[ \boxed{4}. \]