`lim_(theta to pi//2) ((1- sin^(3) theta)/(cos^(2)theta)) =1/2`

Answer

To solve the limit \( \lim_{\theta \to \frac{\pi}{2}} \frac{1 - \sin^3 \theta}{\cos^2 \theta} \), we will follow these steps:

### Step 1: Identify the limit form
As \( \theta \) approaches \( \frac{\pi}{2} \), we notice that:
- \( \sin \theta \) approaches \( 1 \)
- \( \cos \theta \) approaches \( 0 \)

Thus, \( \cos^2 \theta \) approaches \( 0 \) and \( 1 - \sin^3 \theta \) approaches \( 1 - 1 = 0 \). This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Factor the numerator
We can factor the numerator \( 1 - \sin^3 \theta \) using the identity for the difference of cubes:
\[ 1 - \sin^3 \theta = (1 - \sin \theta)(1 + \sin \theta + \sin^2 \theta) \]

### Step 3: Rewrite the limit
Substituting the factorization into the limit, we have:
\[ \lim_{\theta \to \frac{\pi}{2}} \frac{(1 - \sin \theta)(1 + \sin \theta + \sin^2 \theta)}{\cos^2 \theta} \]

### Step 4: Use the identity for \( \cos^2 \theta \)
We know that:
\[ \cos^2 \theta = 1 - \sin^2 \theta = (1 - \sin \theta)(1 + \sin \theta) \]

Thus, we can rewrite the limit as:
\[ \lim_{\theta \to \frac{\pi}{2}} \frac{(1 - \sin \theta)(1 + \sin \theta + \sin^2 \theta)}{(1 - \sin \theta)(1 + \sin \theta)} \]

### Step 5: Cancel common factors
Provided \( \sin \theta \neq 1 \) (which is valid as we approach but do not reach \( \frac{\pi}{2} \)), we can cancel \( (1 - \sin \theta) \):
\[ \lim_{\theta \to \frac{\pi}{2}} \frac{1 + \sin \theta + \sin^2 \theta}{1 + \sin \theta} \]

### Step 6: Substitute the limit
Now, substituting \( \theta = \frac{\pi}{2} \):
- \( \sin \frac{\pi}{2} = 1 \)

Thus, we have:
\[ \frac{1 + 1 + 1^2}{1 + 1} = \frac{3}{2} \]

### Conclusion
The limit evaluates to:
\[ \lim_{\theta \to \frac{\pi}{2}} \frac{1 - \sin^3 \theta}{\cos^2 \theta} = \frac{3}{2} \]

### Final Answer
The statement in the question that the limit equals \( \frac{1}{2} \) is false; the correct limit is \( \frac{3}{2} \).

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