<s>[INST] 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!
 [/INST] Hint
First of all, let $x=\frac 1y$ to make the expression
$$A=\frac{(3 y+1) \log \left(\frac{1+a y}{1+b y}\right)+2 y}{y^2}$$ Now, using Taylor series, you should have
$$A=\frac{a-b+2}{y}+\frac{1}{2} \left(-a^2+6 a+b^2-6 b\right)+\frac{1}{6} y \left(2
   a^3-9 a^2-2 b^3+9 b^2\right)+O\left(y^2\right)$$ So
$$a-b+2=0 \qquad \text{and} \qquad \frac{1}{2} \left(-a^2+6 a+b^2-6 b\right)=2$$ This is easy to solve.
</s>