Let [x] denote the greatest integer less than or equal to Then:
`lim_(x to 0) (tan (pi sin^2x) + (|x|-sin (x[x]))^2)/x^2` :

Options

A. does not exist
B. equals 0
C. equals `pi+1`
D. equals `pi`

Explanation

To solve the limit

\[ \lim_{x \to 0} \frac{\tan(\pi \sin^2 x) + (|x| - \sin(x[x]))^2}{x^2} \]

we will analyze the expression separately for \( x \to 0^+ \) (approaching from the right) and \( x \to 0^- \) (approaching from the left).

### Step 1: Analyze the limit as \( x \to 0^- \)

For \( x < 0 \):
- \( |x| = -x \)
- The greatest integer function \( [x] = -1 \) since \( x \) is between -1 and 0.

Thus, we rewrite the limit:

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

Using the identity \( \sin(-x) = -\sin(x) \), we have:

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

### Step 2: Simplify the expression

Now, we simplify \( (-x + \sin x)^2 \):

\[ (-x + \sin x)^2 = (x - \sin x)^2 \]

Thus, the limit becomes:

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

### Step 3: Evaluate \( \tan(\pi \sin^2 x) \)

Using the small angle approximation \( \sin x \approx x \) as \( x \to 0 \), we have:

\[ \sin^2 x \approx x^2 \]

So,

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

Using the limit \( \tan u \approx u \) as \( u \to 0 \):

\[ \tan(\pi x^2) \approx \pi x^2 \]

### Step 4: Evaluate \( (x - \sin x)^2 \)

Using the Taylor series expansion for \( \sin x \):

\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]

Thus,

\[ x - \sin x \approx \frac{x^3}{6} \]

So,

\[ (x - \sin x)^2 \approx \left(\frac{x^3}{6}\right)^2 = \frac{x^6}{36} \]

### Step 5: Substitute back into the limit

Now we substitute back into the limit:

\[ \lim_{x \to 0^-} \frac{\pi x^2 + \frac{x^6}{36}}{x^2} \]

This simplifies to:

\[ \lim_{x \to 0^-} \left(\pi + \frac{x^4}{36}\right) = \pi \]

### Step 6: Analyze the limit as \( x \to 0^+ \)

For \( x > 0 \):
- \( |x| = x \)
- The greatest integer function \( [x] = 0 \) since \( x \) is between 0 and 1.

Thus, we rewrite the limit:

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

This simplifies to:

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

Using the same approximations as before, we have:

\[ \lim_{x \to 0^+} \frac{\pi x^2 + x^2}{x^2} = \lim_{x \to 0^+} \left(\pi + 1\right) = \pi + 1 \]

### Step 7: Compare the limits

Now we have:

- Left-hand limit as \( x \to 0^- \): \( \pi \)
- Right-hand limit as \( x \to 0^+ \): \( \pi + 1 \)

Since the left-hand limit and right-hand limit are not equal, the overall limit does not exist.

### Final Answer:

\[ \text{The limit does not exist.} \]