How do I solve for $a$ and $b$ in $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$? Given $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$
I need to solve for $a$ and $b$, so here we go,
$\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right)$
$= \lim\limits_{x \to ∞} x \left(2 +(3+x) \left(\ln (1+\frac{a}{x}) - \ln(1+\frac{b}{x}) \right) \right)$
$=\lim\limits_{x \to ∞} x \left(2 +(3+x)\left( \dfrac{a-b}{x} \right) \right)$
$=\lim\limits_{x \to ∞}  \left(2x +(3+x)\left( a-b \right) \right)$
$=\lim\limits_{x \to ∞} \left(2x + 3(a-b) + x(a-b) \right) $
Since the limit exists, $2x$ and $x(a-b)$ must cancel out so $a-b = 2$
But if I do this, I'm left with $3(a-b) = 2$ which gives me a contradiction since $a-b$ is supposed to be $2$.
What do I do now? And where exactly have I gone wrong?
Thank you!

Set $1/x=h$
$$\lim_{h\to0^+}\dfrac{2h+(3h+1)\{\ln(1+ah)-\ln(1+bh)\}}{h^2}$$
$$=3\left(a\lim_{h\to0^+}\dfrac{\ln(1+ah)-1}{ah}-b\lim_{h\to0^+}\dfrac{\ln(1+bh)-1}{bh}\right)+\lim_{h\to0^+}\dfrac{2h+\ln(1+ah)-\ln(1+bh)}{h^2}$$
$$=3(a-b)+\lim_{h\to0^+}\dfrac{2h+\ln(1+ah)-\ln(1+bh)}{h^2}$$
$$=3(a-b)+a^2\lim_{h\to0^+}\dfrac{\ln(1+ah)-ah}{(ah)^2}-b^2\lim_{h\to0^+}\dfrac{\ln(1+bh)-bh}{(bh)^2}+\lim_{h\to0^+}\dfrac{(2+a-b)h}{h^2}$$
$$=3(a-b)+\dfrac{b^2-a^2}2+\lim_{h\to0^+}\dfrac{(2+a-b)h}{h^2}$$ using  Are all limits solvable without L'Hôpital Rule or Series Expansion,
Clearly, we need $2+a-b=0$  and $3(a-b)+\dfrac{b^2-a^2}2=2$
