JEE Advance 2025 Paper 1

For $x > 0$  , let $y_{1}, y_{2}, y_{3}$   satisfy $\frac{dy_{1}}{dx} - y_{1}\sin^{2}x = 0, y_{1}(1)=5$  ; $\frac{dy_{2}}{dx} - y_{2}\cos^{2}x = 0, y_{2}(1)=1/3$  ; $\frac{dy_{3}}{dx} - y_{3}(\frac{2-x^{3}}{x^{3}}) = 0, y_{3}(1)=\frac{3}{5e}$  . Find $\lim_{x \to 0^{+}} \frac{y_{1}y_{2}y_{3} + 2x}{e^{3x}\sin x}$  .