`Lt_(x rarr 0)(int_(0)^(x^(2)) (tan^(-1) t)^(2) dt)/(int_(0)^(x^(2)) (sin sqrtt dt)`=

Options

A. `-1`
B. `-1//2`
C. `0`
D. 1

Answer

To solve the limit problem

\[ \lim_{x \to 0} \frac{\int_{0}^{x^2} (\tan^{-1} t)^2 \, dt}{\int_{0}^{x^2} \sin \sqrt{t} \, dt} \]

we start by substituting \(x = 0\). This gives us:

\[ \frac{\int_{0}^{0} (\tan^{-1} t)^2 \, dt}{\int_{0}^{0} \sin \sqrt{t} \, dt} = \frac{0}{0} \]

This is an indeterminate form, so we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \), we can differentiate the numerator and the denominator.

### Step 1: Differentiate the Numerator and Denominator

Using the Fundamental Theorem of Calculus, we differentiate the numerator and denominator:

1. **Numerator**:
\[ \frac{d}{dx} \left( \int_{0}^{x^2} (\tan^{-1} t)^2 \, dt \right) = (\tan^{-1}(x^2))^2 \cdot \frac{d}{dx}(x^2) = (\tan^{-1}(x^2))^2 \cdot 2x \]

2. **Denominator**:
\[ \frac{d}{dx} \left( \int_{0}^{x^2} \sin \sqrt{t} \, dt \right) = \sin(\sqrt{x^2}) \cdot \frac{d}{dx}(x^2) = \sin(x) \cdot 2x \]

### Step 2: Rewrite the Limit

Now we can rewrite the limit using these derivatives:

\[ \lim_{x \to 0} \frac{(\tan^{-1}(x^2))^2 \cdot 2x}{\sin(x) \cdot 2x} \]

### Step 3: Simplify the Expression

The \(2x\) terms in the numerator and denominator cancel out (as long as \(x \neq 0\)):

\[ \lim_{x \to 0} \frac{(\tan^{-1}(x^2))^2}{\sin(x)} \]

### Step 4: Evaluate the Limit

As \(x \to 0\), we know that \(\tan^{-1}(x^2) \to 0\) and \(\sin(x) \to 0\). We can rewrite \(\tan^{-1}(x^2)\) in terms of \(x\):

\[ \tan^{-1}(x^2) \sim x^2 \quad \text{as } x \to 0 \]

Thus,

\[ (\tan^{-1}(x^2))^2 \sim (x^2)^2 = x^4 \]

So we have:

\[ \lim_{x \to 0} \frac{x^4}{\sin(x)} \]

Using the fact that \(\sin(x) \sim x\) as \(x \to 0\):

\[ \lim_{x \to 0} \frac{x^4}{x} = \lim_{x \to 0} x^3 = 0 \]

### Final Result

Thus, the limit evaluates to:

\[ \boxed{0} \]