$\displaystyle
\frac{\ln{\frac{x+3}{x+1}}}{1/x} = \frac{2x^2}{(x + 1)(x + 3)}
$
This is related to trying to get lim->infinity so technically it is calculus
Hello, TYTY!
The limit goes to $\displaystyle \tfrac{0}{0}$ . . . so we can apply L'Hopital$\displaystyle \lim_{x\to\infty} \frac{\ln{\frac{x+3}{x+1}}}{\frac{1}{x}} \:=\: \lim_{x\to\infty} \frac{2x^2}{(x + 1)(x + 3)}$
We have: .$\displaystyle \frac{\ln (x+3) - \ln(x+1)}{x^{-1}} $
Apply L'Hopital: .$\displaystyle \frac{\frac{1}{x+3}-\frac{1}{x+1}} {-x^{-2}} \;=\;\frac{\frac{-2}{(x+1)(x+3)}}{-\frac{1}{x^2}} \;=\;\frac{2x^2}{(x+1)(x+3)}$