`lim_(xto0)(tan([-pi^(2)]x^(2))-x^(2)tan([-pi^(2)]))/(sin^(2)x)` equals (where [.] denotes the greater ineteger function)

Options

A. 0
B. 1
C. `tan 10 -10`
D. `oo`

Answer

To solve the limit problem

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

we will follow these steps:

### Step 1: Identify the value of \(\lfloor -\pi^2 \rfloor\)

First, we need to find the value of \(\lfloor -\pi^2 \rfloor\). Since \(\pi^2 \approx 9.87\), we have:

\[ -\pi^2 \approx -9.87 \implies \lfloor -\pi^2 \rfloor = -10. \]

### Step 2: Substitute \(\lfloor -\pi^2 \rfloor\) into the limit

Now we substitute \(-10\) into the limit:

\[ \lim_{x \to 0} \frac{\tan(-10 x^2) - x^2 \tan(-10)}{\sin^2 x}. \]

### Step 3: Use the small angle approximation for \(\tan\)

For small values of \(x\), we can use the approximation \(\tan(y) \approx y\) when \(y\) is small. Thus, we have:

\[ \tan(-10 x^2) \approx -10 x^2. \]

### Step 4: Substitute the approximation into the limit

Substituting this back into our limit gives:

\[ \lim_{x \to 0} \frac{-10 x^2 - x^2 \tan(-10)}{\sin^2 x}. \]

### Step 5: Simplify the numerator

The numerator simplifies to:

\[ -10 x^2 - x^2 \tan(-10) = -x^2 (10 + \tan(-10)). \]

### Step 6: Use the limit for \(\sin^2 x\)

As \(x\) approaches \(0\), we can use the fact that \(\sin x \approx x\), thus:

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

### Step 7: Substitute \(\sin^2 x\) into the limit

Now, substituting this into our limit gives:

\[ \lim_{x \to 0} \frac{-x^2 (10 + \tan(-10))}{x^2}. \]

### Step 8: Cancel \(x^2\)

We can cancel \(x^2\) from the numerator and denominator:

\[ \lim_{x \to 0} - (10 + \tan(-10)). \]

### Step 9: Evaluate the limit

Since there are no terms involving \(x\) left, we can directly evaluate the limit:

\[ -(10 + \tan(-10)). \]

### Step 10: Calculate \(\tan(-10)\)

Since \(\tan(-10) = -\tan(10)\), we have:

\[ -(10 - \tan(10)). \]

### Final Result

Thus, the final result is:

\[ \tan(10) - 10. \]