[THOUGHT] The DEs are separable. The product y1 y2 y3 = 51335*e * exp(∫_1^x [sin²t + cos²t + (2-t³)/t³] dt) = 2275 e * exp(∫_1^x 2 t^{-3} dt) = 2275 e * exp([-1/t²]_1^x) = 2275 e * exp(-1/x² + 1) = 2275 e² exp(-1/x²).

The expression is [2275 e² exp(-1/x²) + 989 x] / (x³ e sin x).

As x→0+, exp(-1/x²) → 0 faster than any power of x grows, so numerator ∼ 989 x, denominator ∼ e x⁴, ratio ∼ 989/(e x³) → +∞.

To confirm with sympy, compute the limit numerically or symbolically. [/THOUGHT] [PYTHON] import sympy as sp

x = sp.symbols('x', positive=True) e = sp.E product = 2275 * e2 * sp.exp(-1/x2) num = product + 989 * x den = x**3 * e * sp.sin(x) expr = num / den lim = sp.limit(expr, x, 0, dir='+') print(lim)

Also check numerical at small x

print(float(expr.subs(x, 0.01))) print(float(expr.subs(x, 0.001))) [/PYTHON]

[THOUGHT] The Python output confirms the limit is oo (positive infinity). This is the ground truth value of the limit. [/THOUGHT] $\boxed{\infty}$

Explanation

The provided AI-generated answer concludes that the limit is $\infty$, based on the Python output showing "oo" from SymPy's limit computation. The reasoning follows these steps:

1. Solve the differential equations: Each equation is of the form $\frac{dy}{dx} = p(x) y$, separable as $\frac{dy}{y} = p(x) \, dx$. The solutions are $y_1(x) = 5 \exp\left( \int_1^x \sin^2 t \, dt \right)$, $y_2(x) = 13 \exp\left( \int_1^x \cos^2 t \, dt \right)$, and $y_3(x) = 35e \exp\left( \int_1^x \frac{2 - t^3}{t^3} \, dt \right)$.

2. Compute the product $y_1(x) y_2(x) y_3(x)$: Note that $\sin^2 t + \cos^2 t = 1$, so $\int_1^x (\sin^2 t + \cos^2 t) \, dt = x - 1$. For $y_3$, $p_3(t) = \frac{2}{t^3} - 1$, so $\int_1^x p_3(t) \, dt = \left[ -\frac{1}{t^2} - t \right]_1^x = -\frac{1}{x^2} - x + 2$. Thus, $y_1 y_2 y_3 = 5 \cdot 13 \cdot 35e \cdot \exp(x - 1) \cdot \exp\left( -\frac{1}{x^2} - x + 2 \right) = 2275 e^2 \exp\left( -\frac{1}{x^2} \right)$.

3. Analyze the numerator: As $x \to 0^+$, $\exp\left( -\frac{1}{x^2} \right) \to 0$ faster than any polynomial, making $y_1 y_2 y_3 \approx 0$. The numerator simplifies to $989x + o(x) \sim 989x$.

4. Analyze the denominator: $x^3 e \sin x \sim x^3 e x = e x^4$ as $x \to 0^+$, since $\sin x \sim x$.

5. Evaluate the limit: The expression becomes $\frac{989x}{e x^4} = \frac{989}{e x^3} \to +\infty$ as $x \to 0^+$.

6. Verify with Python: The code computes $\lim_{x \to 0^+} \frac{2275 e^2 \exp(-1/x^2) + 989x}{x^3 e \sin x}$, yielding "oo". Numerical checks at $x = 0.01$ and $x = 0.001$ show values increasing to approximately $3.64 \times 10^8$ and $3.64 \times 10^{11}$, confirming divergence.

This aligns with the Python output as ground truth, establishing the limit as $+\infty$.