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Step-by-Step Solution
Step 1: Identify the differential equation
We are given:
$$f'(x) \;=\; 7 \;-\; \frac{3}{4}\,\frac{f(x)}{x}, \quad x > 0.$$
This can be rewritten in a standard linear differential equation form:
$$\frac{df(x)}{dx} + \frac{3}{4}\,\frac{f(x)}{x} \;=\; 7.$$
Let us set \(y = f(x)\) for convenience in notation, so the equation becomes
$$\frac{dy}{dx} + \frac{3}{4}\,\frac{y}{x} \;=\; 7.$$
Step 2: Determine the integrating factor
The integrating factor (IF) is found by:
$$
\text{IF} \;=\; \exp\!\Bigg(\int \frac{3}{4}\,\frac{1}{x}\,dx\Bigg)
\;=\;\exp\!\bigg(\frac{3}{4} \ln|x|\bigg)
\;=\;|x|^{\tfrac{3}{4}}.
$$
Since \(x > 0\), we can drop the absolute value and write:
$$
\text{IF} \;=\; x^{\tfrac{3}{4}}.
$$
Step 3: Multiply the differential equation by the integrating factor
We multiply each term in
\(\frac{dy}{dx} + \frac{3}{4}\,\frac{y}{x} = 7\)
by \(x^{\tfrac{3}{4}}\):
$$
x^{\tfrac{3}{4}} \frac{dy}{dx}
\;+\;\frac{3}{4} \, x^{\tfrac{3}{4}} \,\frac{y}{x}
\;=\;7\,x^{\tfrac{3}{4}}.
$$
Observe that the left side is the derivative of \(\bigl(y\,x^{\tfrac{3}{4}}\bigr)\). Thus, we get:
$$
\frac{d}{dx}\Bigl(y\,x^{\tfrac{3}{4}}\Bigr) \;=\; 7\,x^{\tfrac{3}{4}}.
$$
Step 4: Integrate to find the general solution
Integrate both sides with respect to \(x\):
$$
y\,x^{\tfrac{3}{4}}
\;=\;
\int 7\,x^{\tfrac{3}{4}}\,dx.
$$
Compute the integral on the right:
$$
\int x^{\tfrac{3}{4}}\,dx
\;=\;
\int x^{0.75}\,dx
\;=\;
\frac{x^{1.75}}{1.75}
\;=\;
\frac{4}{7}\,x^{\tfrac{7}{4}}.
$$
Hence,
$$
y\,x^{\tfrac{3}{4}}
\;=\;
7 \times \frac{4}{7}\,x^{\tfrac{7}{4}}
\;+\; C
\;=\;
4\,x^{\tfrac{7}{4}}
\;+\; C,
$$
where \(C\) is the constant of integration. Therefore,
$$
y \;=\; f(x)
\;=\;
4\,x
\;+\; C\,x^{-\tfrac{3}{4}}.
$$
Step 5: Evaluate the limit as \(x \to 0^+\) of \(x\,f\bigl(\tfrac{1}{x}\bigr)\)
We want
$$
\lim_{x \to 0^+}\bigl[x \,f\bigl(\tfrac{1}{x}\bigr)\bigr].
$$
First substitute \(t = \tfrac{1}{x}\). Then as \(x \to 0^+\), \(t \to +\infty\). So we write:
$$
f\bigl(\tfrac{1}{x}\bigr)
\;=\;
4\left(\tfrac{1}{x}\right)
\;+\;
C\left(\tfrac{1}{x}\right)^{-\tfrac{3}{4}}
\;=\;
\frac{4}{x}
\;+\;
C\,x^{\tfrac{3}{4}}.
$$
Now multiply by \(x\):
$$
x\,f\bigl(\tfrac{1}{x}\bigr)
\;=\;
x\left(\frac{4}{x} + C\,x^{\tfrac{3}{4}}\right)
\;=\;
4 + C\,x^{\tfrac{7}{4}}.
$$
As \(x \to 0^+\), \(x^{\tfrac{7}{4}} \to 0\). Therefore,
$$
\lim_{x \to 0^+} \left(4 + C\,x^{\tfrac{7}{4}}\right)
\;=\;
4.
$$
Conclusion
Hence,
$$
\lim_{x \to 0^+} x\,f\Bigl(\frac{1}{x}\Bigr) \;=\; 4.
$$
