Let f be a differentiable function such that $f^{\mathrm{\prime }}\left(x\right)=7-\frac{3}{4}\frac{f\left(x\right)}{x},(x>0)$ and $f\left(1\right)\ne 4$ Then, $\underset{x\to 0}{lim} xf\left(\frac{1}{x}\right)$

We have,

$\begin{array}{l}{f}^{\mathrm{\prime }}\left(x\right)=7-\frac{3f\left(x\right)}{4}\frac{x}{4}\\ \Rightarrow \frac{d}{dx}\left(f\right(x\left)\right)+\frac{3}{4x}f\left(x\right)=7 ... \left(i\right)\end{array}$

This is a linear differential equation with

I.F.$=e^{\int }\frac{3}{4x}dx=e^{\frac{3}{4}\log x}=x^{\frac{4}{3}}$

Multiplying (i) with $x^{3/4}$ and integrating, we obtain

$\begin{array}{l}f\left(x\right)x^{3/4}=7\int x^{3/4}dx+C\\ \Rightarrow f\left(x\right)x^{3/4}=4x^{3/4}+C\\ \Rightarrow f\left(x\right)=4x+C{x}^{-3/4} ... \left(ii\right)\\ \Rightarrow f\left(\frac{1}{x}\right)=\frac{4}{x}+C{x}^{3/4} ... \left(iii\right)\end{array}$

$\therefore \underset{x\to 0}{lim} xf\left(\frac{1}{x}\right)=\underset{x\to 0}{lim} \left(4+C{x}^{3/4}\right)=4$