Multivariable Equation With Single Variable Limits?!

This relies a bit on the form of the question, but given the limits exist, $y=\alpha x +\beta$ will be the best possible linear approximation to the function for large $x$ . Hence we can write $(\alpha x +\beta)^2 (\alpha x +\beta+1) \approx x(x+1)^2$

This function will be dominated by its cubic term for large $x$ . Comparing coefficients of $x^3$ , we find $\alpha^3=1$ so $\alpha=1$ and our approximation becomes $(x +\beta)^2 (x +\beta+1) \approx x(x+1)^2$

In order to find $\beta$ , we need to go one step further and compare coefficients of $x^2$ : $3\beta+1=2$ so that $\beta=\frac13$ and the required answer is $\boxed{133333}$ .

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Even without assuming the limits exist, it's fairly clear that the function will be asymptotically linear. We can just say again that for large $x$ , only the cubic terms matter, so $y\approx x$ in the limit; alternatively returning to the original equation, $\begin{aligned} y^3+y&=x^3+2x^2+x \\ \frac{y^3}{x^3}+\frac{y}{x^3} &=1+\frac{2}{x}+\frac{1}{x^2} \end{aligned}$

Letting $x \to \infty$ here only the first terms on each side remain; so again $\frac{y^3}{x^3}=1$

Keep in mind that when I write $x$ and $y$ I am referring to very large values of $x$ and $y$ .

From the provided limits, we know that for for a sufficiently large $x$ and $y$ , $y = \alpha x + \beta$ . Thus, $\frac{dy}{dx} = \alpha$ . Rearranging the given equation $y^{2}\left(y+1\right)=x\left(x+1\right)^{2}$ gives us $y^{2}\left(y+1\right)-x\left(x+1\right)^{2}=0$ . Taking the derivative of this equation with respect to $x$ yields $2y\left(y+1\right)\frac{dy}{dx}+y^{2}\frac{dy}{dx}-\left(x+1\right)^{2}-2x\left(x+1\right)=0$ which, after simplifying and substituting $\frac{dy}{dx} = \alpha$ , yields $3\alpha y^{2}+2\alpha y-\left(3x^{2}+4x+1\right)=0$

This is a multivariable quadratic equation which we can use to solve for $y$ . For this problem, we are only expected to solve for positive values of $y$ . Thus, $y=\frac{-2\alpha+\sqrt{4\alpha^{2}+12\alpha\left(3x^{2}+4x+1\right)}}{6\alpha}$ which simplifies to $y=-\frac{1}{3}+\frac{1}{3}\sqrt{1+\frac{1}{\alpha}\left(9x^{2}+12x+3\right)}$

Upon completing the square we get $y=-\frac{1}{3}+\frac{1}{3}\sqrt{1+\frac{1}{\alpha}\left(9\left(x+\frac{2}{3}\right)^{2}-1\right)}$ . For large $x$ , the constants in the expression $\frac{1}{3}\sqrt{1+\frac{1}{\alpha}\left(9\left(x+\frac{2}{3}\right)^{2}-1\right)}$ effectively disappear yielding $y=\sqrt{\frac{1}{\alpha}}x+\left(\frac{2}{3}\sqrt{\frac{1}{\alpha}}-\frac{1}{3}\right)$ . Since $y = \alpha x + \beta$ , $\alpha x+\beta=\sqrt{\frac{1}{\alpha}}x+\left(\frac{2}{3}\sqrt{\frac{1}{\alpha}}-\frac{1}{3}\right)$ meaning $\alpha = \sqrt{\frac{1}{\alpha}}$ and $\beta = \frac{2}{3}\sqrt{\frac{1}{\alpha}}-\frac{1}{3}$ . There is only one solution $(\alpha, \beta)$ that satisfies this system of equations: $(\alpha, \beta) = (1, \frac{1}{3})$ . Thus, $\lfloor 10^{5}\left(\alpha+\beta\right) \rfloor = \boxed{133333}$ .

This approximation for $y$ appears to be pretty accurate at small values of $x$ . For example, for $x = 1$ the error between the true value of $y$ and its approximation is about $0.0187$ (I estimated the true $(x, y$ ) to be $(1, 1.314596213)$ by doing a binary search for $y$ with the condition $y^2(y+1) = 4$ ).