The value of `lim_(x rarr (3pi)/(4)) (4sin^(2) x cosx - cos x + sinx)/(sin x + cos x ) ` is equal to

Options

A. `-1`
B. 0
C. 1
D. none of these

Answer

To find the value of the limit

\[ \lim_{x \to \frac{3\pi}{4}} \frac{4\sin^2 x \cos x - \cos x + \sin x}{\sin x + \cos x}, \]

we will follow these steps:

### Step 1: Evaluate the limit directly
First, we substitute \( x = \frac{3\pi}{4} \) into the expression.

- We know that:
\[ \sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}. \]

Substituting these values into the limit expression:

- The numerator becomes:
\[ 4\left(\frac{1}{\sqrt{2}}\right)^2\left(-\frac{1}{\sqrt{2}}\right) - \left(-\frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}}\right). \]
Simplifying this:
\[ = 4 \cdot \frac{1}{2} \cdot \left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} + \frac{2}{\sqrt{2}} = 0. \]

- The denominator becomes:
\[ \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0. \]

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 \( f(c) = g(c) = 0 \).

We need to differentiate the numerator and denominator.

- **Numerator**: \( f(x) = 4\sin^2 x \cos x - \cos x + \sin x \)
- Differentiate using the product and chain rules:
\[ f'(x) = 4(2\sin x \cos x \cdot \cos x - \sin^2 x \sin x) + \sin x. \]
Simplifying this gives:
\[ = 8\sin x \cos^2 x - 4\sin^3 x + \cos x. \]

- **Denominator**: \( g(x) = \sin x + \cos x \)
- Differentiate:
\[ g'(x) = \cos x - \sin x. \]

### Step 3: Evaluate the limit again
Now we evaluate:

\[ \lim_{x \to \frac{3\pi}{4}} \frac{8\sin x \cos^2 x - 4\sin^3 x + \cos x}{\cos x - \sin x}. \]

Substituting \( x = \frac{3\pi}{4} \):

- The new numerator:
\[ 8\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right)^2 - 4\left(\frac{1}{\sqrt{2}}\right)^3 + \left(-\frac{1}{\sqrt{2}}\right). \]
Simplifying:
\[ = 8 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{2} - 4 \cdot \frac{1}{2\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} - \frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{4 - 2 - 1}{\sqrt{2}} = \frac{1}{\sqrt{2}}. \]

- The new denominator:
\[ -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}. \]

Now we have:

\[ \lim_{x \to \frac{3\pi}{4}} \frac{\frac{1}{\sqrt{2}}}{-\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{-1}{\sqrt{2}} = -\frac{1}{2}. \]

### Final Result
Thus, the value of the limit is:

\[ \boxed{-1}. \]