Find The value of `lim_(x rarr pi) (1+cos^(3)x)/(sin^(2) x)` is

Options

A. `(1)/(3) `
B. `(3)/(2) `
C. `(-1)/(4) `
D. `(-3)/(2)`

Answer

To find the value of the limit

\[ \lim_{x \to \pi} \frac{1 + \cos^3 x}{\sin^2 x}, \]

we will follow these steps:

### Step 1: Direct Substitution
First, we substitute \(x = \pi\) directly into the expression:

- The numerator becomes:
\[ 1 + \cos^3(\pi) = 1 + (-1)^3 = 1 - 1 = 0. \]

- The denominator becomes:
\[ \sin^2(\pi) = 0^2 = 0. \]

Since both the numerator and denominator evaluate to 0, we have an indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \(\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0}\) or \(\frac{\infty}{\infty}\), then:

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

provided that the limit on the right exists.

Let:
- \(f(x) = 1 + \cos^3 x\)
- \(g(x) = \sin^2 x\)

Now, we need to find the derivatives \(f'(x)\) and \(g'(x)\).

### Step 3: Differentiate the Numerator and Denominator
1. Differentiate \(f(x)\):
\[ f'(x) = \frac{d}{dx}(1 + \cos^3 x) = 0 + 3\cos^2 x \cdot (-\sin x) = -3\cos^2 x \sin x. \]

2. Differentiate \(g(x)\):
\[ g'(x) = \frac{d}{dx}(\sin^2 x) = 2\sin x \cos x. \]

### Step 4: Rewrite the Limit
Now we can rewrite the limit using L'Hôpital's Rule:
\[ \lim_{x \to \pi} \frac{f'(x)}{g'(x)} = \lim_{x \to \pi} \frac{-3\cos^2 x \sin x}{2\sin x \cos x}. \]

### Step 5: Simplify the Expression
We can simplify the expression:
\[ = \lim_{x \to \pi} \frac{-3\cos^2 x}{2\cos x} = \lim_{x \to \pi} \frac{-3\cos x}{2}. \]

### Step 6: Substitute \(x = \pi\)
Now we substitute \(x = \pi\):
\[ = \frac{-3\cos(\pi)}{2} = \frac{-3(-1)}{2} = \frac{3}{2}. \]

### Final Answer
Thus, the value of the limit is
\[ \frac{3}{2}. \]