Let $$y = f(x)$$ be the solution of the differential equation $$y(x+1)dx - x^2 dy = 0$$, $$y(1) = e$$. Then $$\lim_{x \to 0^+} f(x)$$ is equal to

To find the limit, we first solve the given differential equation by separating the variables.

The differential equation is given as:

$$y(x + 1) \, dx - x^2 \, dy = 0$$

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Step 1: Separate the variables and integrate

Rearranging the terms to group $$x$$ and $$y$$ variables on opposite sides:

$$x^2 \, dy = y(x + 1) \, dx$$

$$\frac{1}{y} \, dy = \frac{x + 1}{x^2} \, dx$$

$$\frac{1}{y} \, dy = \left(\frac{1}{x} + \frac{1}{x^2}\right) \, dx$$

Integrating both sides:

$$\int \frac{1}{y} \, dy = \int \left(\frac{1}{x} + x^{-2}\right) \, dx$$

$$\log_e|y| = \log_e|x| - \frac{1}{x} + C$$

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Step 2: Find the constant of integration $$C$$ using the initial condition

We are given that $$y(1) = e$$. Substituting $$x = 1$$ and $$y = e$$ into the integrated equation:

$$\log_e(e) = \log_e(1) - \frac{1}{1} + C$$

$$1 = 0 - 1 + C \implies C = 2$$

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Step 3: Express $$y = f(x)$$ explicitly

Substituting $$C = 2$$ back into the general solution equation:

$$\log_e y = \log_e x - \frac{1}{x} + 2$$

Taking the exponential on both sides to isolate $$y$$:

$$y = e^{\log_e x - \frac{1}{x} + 2}$$

$$f(x) = e^{\log_e x} \cdot e^{2 - \frac{1}{x}} = x \cdot e^{2 - \frac{1}{x}}$$

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Step 4: Evaluate the limit as $$x \to 0^+$$

$$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left(x \cdot e^{2 - \frac{1}{x}}\right)$$

As $$x \to 0^+$$, the component terms behave as follows:

$$x \to 0$$

$$\frac{1}{x} \to \infty \implies -\frac{1}{x} \to -\infty$$

$$e^{2 - \frac{1}{x}} \to e^{-\infty} = 0$$

Evaluating the product limit directly:

$$\lim_{x \to 0^+} f(x) = 0 \times 0 = 0$$

Therefore, the value of the limit is equal to 0.