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

Options

A. [2, 5)
B. (1, 5)
C. (-1, 5)
D. `(-oo, oo)`

Explanation

To find the value of the limit

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

we start by substituting \(x = \pi\):

1. **Substitution**:
\[ \cos(\pi) = -1 \quad \text{and} \quad \sin(\pi) = 0. \]
Therefore, we have:
\[ 1 + \cos^3(\pi) = 1 + (-1)^3 = 1 - 1 = 0, \]
and
\[ \sin^2(\pi) = 0^2 = 0. \]
This gives us the indeterminate form \(\frac{0}{0}\).

**Hint**: When you encounter a \(\frac{0}{0}\) form, consider using L'Hôpital's Rule or algebraic manipulation to simplify the expression.

2. **Applying L'Hôpital's Rule**:
Since we have the indeterminate form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \quad \text{if the limit exists.} \]
Here, let \(f(x) = 1 + \cos^3 x\) and \(g(x) = \sin^2 x\).

3. **Finding the derivatives**:
- The derivative of \(f(x)\):
\[ f'(x) = 0 + 3\cos^2 x (-\sin x) = -3\cos^2 x \sin x. \]
- The derivative of \(g(x)\):
\[ g'(x) = 2\sin x \cos x. \]

4. **Applying L'Hôpital's Rule**:
Now we can rewrite the limit:
\[ \lim_{x \to \pi} \frac{f'(x)}{g'(x)} = \lim_{x \to \pi} \frac{-3\cos^2 x \sin x}{2\sin x \cos x}. \]
We can simplify this:
\[ = \lim_{x \to \pi} \frac{-3\cos^2 x}{2\cos x} = \lim_{x \to \pi} \frac{-3\cos x}{2}. \]

5. **Substituting \(x = \pi\)**:
Now substitute \(x = \pi\):
\[ \lim_{x \to \pi} \frac{-3\cos(\pi)}{2} = \frac{-3(-1)}{2} = \frac{3}{2}. \]

Thus, the value of the limit is

\[ \frac{3}{2}. \]