Question:

Step 1: To find $\displaystyle\lim_{x\to0^+}f(x)$, we first need to solve the given differential equation $y(x+1)\,dx-x^2\,dy=0$. This equation can be rearranged to separate the variables, which will allow us to integrate and find the solution $y=f(x)$. Step 2: Rearranging the differential equation, we get $y(x+1)\,dx = x^2\,dy$. Separating the variables yields $\frac{dy}{y} = \frac{x+1}{x^2}\,dx$. This step is crucial as it sets up the equation for integration. Step 3: Now, we integrate both sides of the equation $\frac{dy}{y} = \frac{x+1}{x^2}\,dx$. The left side integrates to $\ln|y|$ and the right side can be integrated by recognizing it as $\frac{x+1}{x^2} = \frac{1}{x} + \frac{1}{x^2}$, which integrates to $\ln|x| - \frac{1}{x} + C$, where $C$ is the constant of integration. Step 4: After integration, we have $\ln|y| = \ln|x| - \frac{1}{x} + C$. To solve for $y$, we exponentiate both sides, resulting in $|y| = e^{\ln|x| - \frac{1}{x} + C} = e^{\ln|x|}e^{-\frac{1}{x}}e^{C} = |x|e^{-\frac{1}{x}}e^{C}$. Since $y$ must be positive due to the initial condition $y(1) = e$, we can drop the absolute value, leading to $y = xe^{-\frac{1}{x}}e^{C}$. Step 5: We use the initial condition $y(1) = e$ to solve for $C$. Substituting $x=1$ and $y=e$ into the equation $y = xe^{-\frac{1}{x}}e^{C}$ gives $e = e^{-1}e^{C}$. Solving for $C$ yields $e^{C} = e^2$, hence $C = 2$. Step 6: Substituting $C = 2$ back into the equation for $y$ gives $y = xe^{-\frac{1}{x}}e^{2}$. Now, we need to find $\displaystyle\lim_{x\to0^+}f(x) = \displaystyle\lim_{x\to0^+}xe^{-\frac{1}{x}}e^{2}$. Step 7: To evaluate $\displaystyle\lim_{x\to0^+}xe^{-\frac{1}{x}}e^{2}$, notice that as $x$ approaches $0$ from the right, $e^{-\frac{1}{x}}$ approaches $0$ much faster than $x$ approaches $0$. However, the limit is of the form $0 \cdot 0$, so we can use L'Hôpital's rule or recognize the behavior of $xe^{-\frac{1}{x}}$ as $x$ approaches $0$. Since $e^{2}$ is a constant factor, it does not affect the limit's behavior as $x$ approaches $0$. Step 8: Recognizing that $\displaystyle\lim_{x\to0^+}xe^{-\frac{1}{x}} = 0$ because the exponential decay of $e^{-\frac{1}{x}}$ dominates the linear approach of $x$ to $0$, we conclude that $\displaystyle\lim_{x\to0^+}f(x) = \displaystyle\lim_{x\to0^+}xe^{-\frac{1}{x}}e^{2} = 0 \cdot e^{2} = 0$. Step 9: Therefore, the final answer is $\boxed{0}$, which corresponds to Option 1.