To solve the limit
\[
\lim_{x \to 0} \frac{\int_{0}^{x^2} \tan^{-1}(t) \, dt}{\int_{0}^{x^2} \sin(\sqrt{t}) \, dt},
\]
we first need to evaluate the behavior of both the numerator and the denominator as \( x \) approaches 0.
### Step 1: Evaluate the limits of the numerator and denominator
As \( x \to 0 \), both \( \int_{0}^{x^2} \tan^{-1}(t) \, dt \) and \( \int_{0}^{x^2} \sin(\sqrt{t}) \, dt \) approach 0. Thus, we have an indeterminate form \( \frac{0}{0} \).
### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:
\[
\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}
\]
if the limit on the right-hand side exists. We differentiate the numerator and denominator with respect to \( x \).
### Step 3: Differentiate the numerator
Using the Fundamental Theorem of Calculus and the chain rule, we differentiate the numerator:
\[
\frac{d}{dx} \left( \int_{0}^{x^2} \tan^{-1}(t) \, dt \right) = \tan^{-1}(x^2) \cdot \frac{d}{dx}(x^2) = \tan^{-1}(x^2) \cdot 2x.
\]
### Step 4: Differentiate the denominator
Similarly, we differentiate the 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 5: Rewrite the limit
Now we can rewrite the limit using the derivatives:
\[
\lim_{x \to 0} \frac{\tan^{-1}(x^2) \cdot 2x}{\sin(x) \cdot 2x}.
\]
We can cancel \( 2x \) from the numerator and denominator (as long as \( x \neq 0 \)):
\[
\lim_{x \to 0} \frac{\tan^{-1}(x^2)}{\sin(x)}.
\]
### Step 6: Evaluate the new limit
As \( x \to 0 \), \( \tan^{-1}(x^2) \) approaches \( \tan^{-1}(0) = 0 \) and \( \sin(x) \) approaches \( \sin(0) = 0 \), which is again an indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule again.
### Step 7: Differentiate again
Differentiate the numerator and denominator again:
1. The derivative of \( \tan^{-1}(x^2) \) is \( \frac{2x}{1 + (x^2)^2} = \frac{2x}{1 + x^4} \).
2. The derivative of \( \sin(x) \) is \( \cos(x) \).
Thus, we have:
\[
\lim_{x \to 0} \frac{\frac{2x}{1 + x^4}}{\cos(x)}.
\]
### Step 8: Evaluate the limit
Now substituting \( x = 0 \):
\[
\frac{2 \cdot 0}{1 + 0^4} \cdot \frac{1}{\cos(0)} = \frac{0}{1} = 0.
\]
### Conclusion
Therefore, the final result is:
\[
\lim_{x \to 0} \frac{\int_{0}^{x^2} \tan^{-1}(t) \, dt}{\int_{0}^{x^2} \sin(\sqrt{t}) \, dt} = 0.
\]
