`lim _(xto0) (sin (pi cos ^(2) (tan (sin x ))))/(x ^(2))=`

Options

A. `pi`
B. `pi/4`
C. `pi/2`
D. none of these

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{\sin(\pi \cos^2(\tan(\sin x)))}{x^2}, \]

we will proceed step by step.

### Step 1: Analyze the limit
As \( x \to 0 \), we notice that both the numerator and denominator approach 0. Specifically, \(\sin(0) = 0\) and \(x^2 \to 0\). This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Rewrite the expression
We can rewrite the limit as:

\[ \lim_{x \to 0} \frac{\sin(\pi \cos^2(\tan(\sin x)))}{x^2}. \]

### Step 3: Use Taylor series expansion
For small values of \(x\), we can use the Taylor series expansion for \(\sin x\) and \(\tan x\):

1. \(\sin x \approx x\)
2. \(\tan x \approx x\)

Thus, we have:

\[ \tan(\sin x) \approx \tan(x) \approx x. \]

### Step 4: Substitute in the limit
Now, substituting \(\tan(\sin x) \approx x\) into \(\cos^2(\tan(\sin x))\):

\[ \cos^2(\tan(\sin x)) \approx \cos^2(x). \]

Using the Taylor expansion for \(\cos x\):

\[ \cos x \approx 1 - \frac{x^2}{2} \implies \cos^2(x) \approx \left(1 - \frac{x^2}{2}\right)^2 \approx 1 - x^2 + O(x^4). \]

### Step 5: Substitute back into the limit
Now we substitute this back into our limit:

\[ \sin(\pi \cos^2(\tan(\sin x))) \approx \sin(\pi (1 - x^2)). \]

Using the identity \(\sin(\pi - x) = \sin x\):

\[ \sin(\pi (1 - x^2)) = \sin(\pi - \pi x^2) = \sin(\pi x^2). \]

### Step 6: Use the small angle approximation
For small angles, \(\sin y \approx y\):

\[ \sin(\pi x^2) \approx \pi x^2. \]

### Step 7: Substitute into the limit
Now substituting this back into our limit:

\[ \lim_{x \to 0} \frac{\pi x^2}{x^2} = \pi. \]

### Conclusion
Thus, the final result is:

\[ \lim_{x \to 0} \frac{\sin(\pi \cos^2(\tan(\sin x)))}{x^2} = \pi. \]

### Final Answer
The limit evaluates to:

\[ \pi. \]