Functional Equation or Limit?

First, i didn't know that primitive means anti-derivative, i had to google,

Clearly, $F'(x) = f(x)$ .

Now, differentiate both sides of the given equation,

$2(f(x)-f'(x)) = 2f(x)f'(x)$ ,

Rearranging the terms,

$f'(x) = \frac{f(x)}{f(x) + 1}$ .

Note that $f'(x) > 0 \forall x \in R$ , hence the value $+\infty$ will be approached as $x$ approaches $+\infty$ . Hence , we can apply L-hopital's rule on $\displaystyle \lim_{x \to \infty} \frac{f(x)}{x}$ .

$\therefore \displaystyle \lim_{x \to \infty} \frac{f(x)}{x} = \lim_{x \to \infty} f'(x) = \lim_{x \to \infty} \frac{f(x)}{f(x)+1} = \lim_{x \to 0} \frac{f'(x)}{f'(x)} = \boxed{1}$

We differentiate the given equation implicitly: $\begin{aligned} 2(f(x) - f'(x)) &= 2f(x)f'(x) \\ f(x)-f'(x) &= f(x)f'(x) \\ f(x) &= (f(x)+1)f'(x) \\ f'(x) &= \dfrac{f(x)}{f(x)+1}. \end{aligned}$ Replacing $f'(x)$ with $\dfrac{dy}{dx}$ and $f(x)$ with $y,$ this becomes the separable differential equation $\qquad \dfrac{dy}{dx} = \dfrac{y}{y+1}. \qquad \text{(1)}$ We separate to get $\dfrac{y+1}{y}dy = dx,$ or $\left(1+\dfrac1y\right)dy = dx.$ Integrating both sides gives

$y + \ln y = x + c.$ (Since $y > 0,$ we can ignore the absolute value signs in $\ln \!\mid \!y\!\mid.$ ) Thus, we have $\dfrac{y+\ln y}{x} = 1 + \dfrac cx,$ so $\qquad \displaystyle \lim_{x \to \infty} \dfrac{y+\ln y}{x} = \displaystyle \lim_{x \to \infty} \left(1 + \dfrac cx\right) = 1. \qquad \text{(2)}$

Now, we prove that $\displaystyle\lim_{x\to\infty}y = \infty,$ using equation $\text{(2)}.$ Assume to the contrary that $\displaystyle\lim_{x\to\infty} y$ is finite. (Remember that $y$ is a function of $x.$ ) Then, $\displaystyle\lim_{x\to\infty}\ln y$ is also finite, since $\ln y < y$ for all $y.$ But then we get $\displaystyle \lim_{x \to \infty} \dfrac{y+\ln y}{x} = 0,$ contradicting equation $\text{(2)}.$ (This is because the numerator approaches a finite number while the denominator approaches $\infty.$ ) Thus, by contradiction, $\lim_{x\to\infty} y = \infty.$

This means that we can apply L'Hospital's Rule on the requested limit: $L = \lim_{x \to \infty} \dfrac yx = \lim_{x\to\infty} \dfrac{dy/dx}{1} = \lim_{x\to\infty} \dfrac{dy}{dx}.$ But from equation $\text{(1)},$ $\dfrac{dy}{dx} = \dfrac{y}{y+1}.$ Since $y$ approaches $\infty$ as $x$ approaches $\infty,$ we conclude that $L = \lim_{x\to\infty} \dfrac{dy}{dx} = \lim_{\boxed{x}\to\infty} \dfrac{y}{y+1} = \lim_{\boxed{y}\to\infty} \dfrac{y}{y+1} = \boxed{1}.$

we can see that : $2F(x)=f^2(x)+2f(x)>0$ . So we can write : $f(x)=\frac{\mathrm{d}F(x)}{\mathrm{d}x}= -1+\sqrt{1+2F(x)}.$ By separation of variables : $C+f(x) +\log(f(x))=\int \frac{\mathrm{d}F}{-1+\sqrt{1+2F(x)}} = \int \ \mathrm{d}x=x.\ \ \ \ \ \ (1)$ Now, it is obvious that $f(x)\to \infty$ as $x\to \infty$ . Therefore : $\lim_{x\to \infty} \frac{\log(f(x))}{f(x)} = 0.$ Back to $(1)$ : $1=\lim_{x\to \infty} \frac{C}{f(x)} + 1+ \frac{\log(f(x))}{f(x)} =\lim_{x\to \infty} \frac{x}{f(x)}.$ This means that : $\lim_{x\to \infty} \frac{f(x)}{x} =\boxed{1}.$