<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] 
Since the limit exists, $2x$ and $x(a−b)$ must cancel out so $a−b=2$.

You wanted to say $2+a-b=0 \Rightarrow \color{red}{a-b=-2}$. 
And you cannot go from the difference of logarithms (line $2$) to the difference of arguments (line $3$), because it is not a factor, but a sum $(2+\cdots)$, which is multiplied by $x$. In other words, you can not apply the asymptotics to an addend, but must apply to a factor.
Alternatively:
$$\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2 \Rightarrow \\
\lim\limits_{x \to ∞} \frac{2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right)}{\frac 1x} = 2 \ \ \ (*)$$
For the limit to exist:
$$\lim\limits_{x \to ∞} \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 0 \Rightarrow \cdots \Rightarrow  a-b=-2.$$
Now use L'Hospital's rule for $(*)$ twice:
$$\lim\limits_{x \to ∞} \frac{\ln (x+a) +\frac{3+x}{x+a}- \ln (x+b) -\frac{3+x}{x+b}}{-\frac 1{x^2}} = \\
=\lim\limits_{x \to ∞} \frac{\frac1{x+a}+\frac{a-3}{(x+a)^2}-\frac1{x+b}-\frac{b-3}{(x+b)^2}}{\frac 2{x^3}} = \\
=\lim_{x\to\infty} -\frac{x^3 (a - b) (2 a b + a x - 3 a + b x - 3 b - 6 x)}{2 (x+a)^2 (x+b)^2}=2 \Rightarrow \\
\color{red}{a+b-6=2}.$$
Hence:
$$\begin{cases}a-b=-2 \\ a+b-6=2 \end{cases} \Rightarrow a=3, b=5.$$
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