Question:

Step 1: Check the indeterminate form of the given limit. Substitute $x=0$ into the numerator and denominator to determine the form of the limit. The numerator becomes: $$ \int_0^{0^2} \sec^2 t\, dt = \int_0^0 \sec^2 t\, dt = 0 $$ The denominator becomes: $$ 0 \cdot \sin 0 = 0 \cdot 0 = 0 $$ Since both the numerator and denominator approach $0$, the limit is of the indeterminate form $\frac{0}{0}$. Step 2: Apply L'Hôpital's rule by differentiating the numerator. Since the limit is of the indeterminate form $\frac{0}{0}$, we apply L'Hôpital's rule. We differentiate the numerator with respect to $x$ using the Fundamental Theorem of Calculus (Part I) and the chain rule. Let $N(x) = \int_0^{x^2} \sec^2 t\, dt$. Then, the derivative of the numerator is: $$ N'(x) = \sec^2(x^2) \cdot \frac{d}{dx}(x^2) = \sec^2(x^2) \cdot 2x $$ Step 3: Apply L'Hôpital's rule by differentiating the denominator. Next, we differentiate the denominator with respect to $x$ using the product rule. Let $D(x) = x \sin x$. Then, the derivative of the denominator is: $$ D'(x) = \frac{d}{dx}(x) \cdot \sin x + x \cdot \frac{d}{dx}(\sin x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x $$ Step 4: Formulate the new limit expression and check its form. Substitute the differentiated numerator and denominator back into the limit expression. The limit becomes: $$ \lim_{x \to 0} \frac{2x \sec^2(x^2)}{\sin x + x \cos x} $$ Now, substitute $x=0$ into this new expression: Numerator: $2(0) \sec^2(0^2) = 0 \cdot \sec^2(0) = 0 \cdot 1 = 0$ Denominator: $\sin 0 + 0 \cos 0 = 0 + 0 = 0$ The limit is still of the indeterminate form $\frac{0}{0}$. Step 5: Apply L'Hôpital's rule a second time by differentiating the new numerator. Since the limit is still of the indeterminate form $\frac{0}{0}$, we apply L'Hôpital's rule again. We differentiate the new numerator, $2x \sec^2(x^2)$, with respect to $x$ using the product rule and chain rule. Let $N_1(x) = 2x \sec^2(x^2)$. $$ N_1'(x) = \frac{d}{dx}(2x) \cdot \sec^2(x^2) + 2x \cdot \frac{d}{dx}(\sec^2(x^2)) $$ $$ N_1'(x) = 2 \sec^2(x^2) + 2x \cdot \left( 2 \sec(x^2) \cdot \frac{d}{dx}(\sec(x^2)) \right) $$ $$ N_1'(x) = 2 \sec^2(x^2) + 4x \sec(x^2) \cdot \left( \sec(x^2) \tan(x^2) \cdot \frac{d}{dx}(x^2) \right) $$ $$ N_1'(x) = 2 \sec^2(x^2) + 4x \sec^2(x^2) \tan(x^2) \cdot 2x $$ $$ N_1'(x) = 2 \sec^2(x^2) + 8x^2 \sec^2(x^2) \tan(x^2) $$ Step 6: Apply L'Hôpital's rule a second time by differentiating the new denominator. Next, we differentiate the new denominator, $\sin x + x \cos x$, with respect to $x$ using the sum rule and product rule. Let $D_1(x) = \sin x + x \cos x$. $$ D_1'(x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(x \cos x) $$ $$ D_1'(x) = \cos x + \left( \frac{d}{dx}(x) \cdot \cos x + x \cdot \frac{d}{dx}(\cos x) \right) $$ $$ D_1'(x) = \cos x + (1 \cdot \cos x + x \cdot (-\sin x)) $$ $$ D_1'(x) = \cos x + \cos x - x \sin x $$ $$ D_1'(x) = 2 \cos x - x \sin x $$ Step 7: Evaluate the limit after the second application of L'Hôpital's rule. Substitute the twice-differentiated numerator and denominator back into the limit expression. The limit becomes: $$ \lim_{x \to 0} \frac{2 \sec^2(x^2) + 8x^2 \sec^2(x^2) \tan(x^2)}{2 \cos x - x \sin x} $$ Now, substitute $x=0$ into this expression: Numerator: $2 \sec^2(0^2) + 8(0)^2 \sec^2(0^2) \tan(0^2) = 2(1)^2 + 8(0)(1)^2(0) = 2 + 0 = 2$ Denominator: $2 \cos(0) - 0 \sin(0) = 2(1) - 0 = 2$ The value of the limit is $\frac{2}{2} = 1$. The final answer is $\boxed{1}$.