A Small Problem From JEE Mains 2019!

$\begin{aligned} f'(x) & = 7 - \frac {3f(x)}{4x} \\ \frac d{dx} f(x) + \frac {{\color{#3D99F6}3}f(x)}{\color{#3D99F6}4x} & = 7 & \small \color{#3D99F6} \text{Multiply both sides by }\mu(x) = e^{\int \frac 3{4x} dx} = x^\frac 34 \\ x^\frac 34 \frac d{dx}f(x) + {\color{#3D99F6}\frac 34 x^{-\frac 14}} f(x) & = 7x^\frac 34 & \small \color{#3D99F6} \text{Since }\frac d{dx} x^\frac 34 = \frac 34 x^{-\frac 14} \\ x^\frac 34 \frac d{dx} f(x) + f(x) \color{#3D99F6}\frac d{dx} x^\frac 34 & = 7x^\frac 34 \\ \frac d{dx}\left( x^\frac 34 f(x)\right) & = 7 x^\frac 34 \\ f(x) & = \frac 1{x^\frac 34}\int 7 x^\frac 34 dx \\ \implies f(x) & = 4x + {\color{#3D99F6}C}x^{-\frac 34} & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \end{aligned}$

$\begin{aligned} \implies \lim_{x \to 0^+} x f\left(\frac 1x\right) & = \lim_{x \to 0^+} x \left(\frac 4x + Cx^\frac 34 \right) \\ & = \lim_{x \to 0^+} 4 + Cx^\frac 74 \\ & = \boxed 4 \end{aligned}$