Find `lim_(thetararr0) (tan(pi*cos^2theta))/(sin(2pisin^2theta))`

Answer

To solve the limit \( \lim_{\theta \to 0} \frac{\tan(\pi \cos^2 \theta)}{\sin(2\pi \sin^2 \theta)} \), we will follow these steps:

### Step 1: Evaluate the limit directly
First, we substitute \( \theta = 0 \):
- \( \cos^2(0) = 1 \) so \( \tan(\pi \cos^2(0)) = \tan(\pi) = 0 \)
- \( \sin^2(0) = 0 \) so \( \sin(2\pi \sin^2(0)) = \sin(0) = 0 \)

This gives us the form \( \frac{0}{0} \), which is indeterminate.

**Hint:** When you get a \( \frac{0}{0} \) form, you can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Since we have \( \frac{0}{0} \), we differentiate the numerator and the denominator with respect to \( \theta \).

**Numerator:**
\[ \frac{d}{d\theta}[\tan(\pi \cos^2 \theta)] = \sec^2(\pi \cos^2 \theta) \cdot \frac{d}{d\theta}[\pi \cos^2 \theta] \]
Using the chain rule:
\[ \frac{d}{d\theta}[\pi \cos^2 \theta] = \pi \cdot 2\cos \theta(-\sin \theta) = -2\pi \cos \theta \sin \theta \]
Thus, the derivative of the numerator is:
\[ \sec^2(\pi \cos^2 \theta) \cdot (-2\pi \cos \theta \sin \theta) \]

**Denominator:**
\[ \frac{d}{d\theta}[\sin(2\pi \sin^2 \theta)] = \cos(2\pi \sin^2 \theta) \cdot \frac{d}{d\theta}[2\pi \sin^2 \theta] \]
Using the chain rule:
\[ \frac{d}{d\theta}[2\pi \sin^2 \theta] = 2\pi \cdot 2\sin \theta \cos \theta = 4\pi \sin \theta \cos \theta \]
Thus, the derivative of the denominator is:
\[ \cos(2\pi \sin^2 \theta) \cdot (4\pi \sin \theta \cos \theta) \]

### Step 3: Rewrite the limit
Now, we can rewrite the limit using these derivatives:
\[ \lim_{\theta \to 0} \frac{-2\pi \cos \theta \sin \theta \sec^2(\pi \cos^2 \theta)}{4\pi \sin \theta \cos \theta \cos(2\pi \sin^2 \theta)} \]

### Step 4: Simplify the expression
Notice that \( \sin \theta \) and \( \cos \theta \) cancel out:
\[ \lim_{\theta \to 0} \frac{-2 \sec^2(\pi \cos^2 \theta)}{4 \cos(2\pi \sin^2 \theta)} = \lim_{\theta \to 0} \frac{-\sec^2(\pi \cos^2 \theta)}{2 \cos(2\pi \sin^2 \theta)} \]

### Step 5: Evaluate the limit
As \( \theta \to 0 \):
- \( \sec^2(\pi \cos^2(0)) = \sec^2(\pi) = \sec^2(1) \)
- \( \cos(2\pi \sin^2(0)) = \cos(0) = 1 \)

Thus, the limit becomes:
\[ \frac{-\sec^2(\pi)}{2 \cdot 1} = \frac{-1}{2} \]

### Final Answer
Therefore, the limit is:
\[ \lim_{\theta \to 0} \frac{\tan(\pi \cos^2 \theta)}{\sin(2\pi \sin^2 \theta)} = -\frac{1}{2} \]