Question

$\lim_{x\to0} \frac{\int_{0}^{x^2} \cos t^2 dt}{x \sin x}$ is equal to

Answer 1

1

Answer 2

The given limit is $\lim_{x\to0} \frac{\int_{0}^{x^2} \cos t^2 dt}{x \sin x}$.

Step 1: Check the form of the limit. As $x \to 0$: Numerator: $\int_{0}^{x^2} \cos t^2 dt \to \int_{0}^{0} \cos t^2 dt = 0$. Denominator: $x \sin x \to 0 \cdot \sin 0 = 0 \cdot 0 = 0$. The limit is of the indeterminate form $\frac{0}{0}$. Therefore, we can apply L'Hopital's Rule.

Step 2: Apply L'Hopital's Rule for the first time. Let $N(x) = \int_{0}^{x^2} \cos t^2 dt$ and $D(x) = x \sin x$. We need to find $N'(x)$ and $D'(x)$.

For $N'(x)$, we use the Leibniz integral rule: $\frac{d}{dx} \int_{a(x)}^{b(x)} h(t) dt = h(b(x)) b'(x) - h(a(x)) a'(x)$. Here, $h(t) = \cos t^2$, $a(x) = 0$, and $b(x) = x^2$. $N'(x) = \cos((x^2)^2) \cdot \frac{d}{dx}(x^2) - \cos(0^2) \cdot \frac{d}{dx}(0)$ $N'(x) = \cos(x^4) \cdot (2x) - \cos(0) \cdot 0$ $N'(x) = 2x \cos(x^4)$

For $D'(x)$, we use the product rule: $\frac{d}{dx}(uv) = u'v + uv'$. $D'(x) = \frac{d}{dx}(x \sin x) = (1)\sin x + x(\cos x)$ $D'(x) = \sin x + x \cos x$

Applying L'Hopital's Rule, the limit becomes: $\lim_{x\to0} \frac{2x \cos(x^4)}{\sin x + x \cos x}$

Step 3: Check the form of the new limit. As $x \to 0$: Numerator: $2(0) \cos(0^4) = 0 \cdot 1 = 0$. Denominator: $\sin 0 + 0 \cos 0 = 0 + 0 = 0$. The limit is still of the indeterminate form $\frac{0}{0}$. So, we apply L'Hopital's Rule again.

Step 4: Apply L'Hopital's Rule for the second time. Let $N_1(x) = 2x \cos(x^4)$ and $D_1(x) = \sin x + x \cos x$. We need to find $N_1'(x)$ and $D_1'(x)$.

For $N_1'(x)$, use the product rule: $N_1'(x) = \frac{d}{dx}(2x \cos(x^4)) = 2 \cdot \cos(x^4) + 2x \cdot (-\sin(x^4) \cdot \frac{d}{dx}(x^4))$ $N_1'(x) = 2 \cos(x^4) - 2x \sin(x^4) \cdot (4x^3)$ $N_1'(x) = 2 \cos(x^4) - 8x^4 \sin(x^4)$

For $D_1'(x)$: $D_1'(x) = \frac{d}{dx}(\sin x + x \cos x) = \cos x + (1 \cdot \cos x + x \cdot (-\sin x))$ $D_1'(x) = \cos x + \cos x - x \sin x$ $D_1'(x) = 2 \cos x - x \sin x$

Applying L'Hopital's Rule again, the limit becomes: $\lim_{x\to0} \frac{2 \cos(x^4) - 8x^4 \sin(x^4)}{2 \cos x - x \sin x}$

Step 5: Evaluate the final limit. As $x \to 0$: Numerator: $2 \cos(0^4) - 8(0)^4 \sin(0^4) = 2 \cos(0) - 0 = 2(1) - 0 = 2$. Denominator: $2 \cos 0 - 0 \sin 0 = 2(1) - 0 = 2$. The limit evaluates to $\frac{2}{2} = 1$.

The final answer is $\boxed{1}$.

Explanation of the solution: The limit is of the form $\frac{0}{0}$. Apply L'Hopital's Rule twice.

1. Differentiate the numerator using Leibniz rule: $\frac{d}{dx} \int_{0}^{x^2} \cos t^2 dt = 2x \cos(x^4)$.
2. Differentiate the denominator using product rule: $\frac{d}{dx} (x \sin x) = \sin x + x \cos x$. The limit becomes $\lim_{x\to0} \frac{2x \cos(x^4)}{\sin x + x \cos x}$, which is still $\frac{0}{0}$.
3. Differentiate the new numerator: $\frac{d}{dx} (2x \cos(x^4)) = 2 \cos(x^4) - 8x^4 \sin(x^4)$.
4. Differentiate the new denominator: $\frac{d}{dx} (\sin x + x \cos x) = 2 \cos x - x \sin x$. The limit becomes $\lim_{x\to0} \frac{2 \cos(x^4) - 8x^4 \sin(x^4)}{2 \cos x - x \sin x}$.
5. Substitute $x=0$: $\frac{2 \cos(0) - 0}{2 \cos(0) - 0} = \frac{2}{2} = 1$.