<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] Your mistake lies in the stp where you replace the difference between logs with the ratio $(a-b) /x$. You can not replace an expression $A$ by another expression $B$ unless $A=B$. While evaluating limit you may find replacements of $A$ by $B$ even when $A\neq B$ but such replacements always represent some hidden steps which the the author assumes to be obvious (an example of such a replacement is in my solution where the factor $x$ is replaced by $x+b$ and it is based on the fact that $x/(x+b) \to 1$ as $x\to\infty$). Also see this answer for details about such replacements. 

The given limit condition implies that $(x+3)\log((x+a)/(x+b))\to - 2$ Then clearly $a\neq b$ and we have $$(a-b)\cdot\frac{x+3}{x+b}\cdot\dfrac{\log\left(1+\dfrac{a-b}{x+b}\right)}{\dfrac{a-b}{x+b}}\to - 2$$ Thus $a-b=-2$ or $a=b-2$. Using this value of $a$ in original limit we see that $$x\left(2+(x+3)\log\left(1-\frac{2}{x+b}\right)\right)\to 2$$ Replacing the first factor by $x+b$ and putting $x+b=2/t$ we see that $$\lim_{t\to 0^{+}}\frac{1}{t}\left(2+\left(\frac{2}{t}+3-b\right)\log(1-t)\right)=1$$ or $$\frac{2t+2\log(1-t)}{t^2}+(3-b)\frac{\log(1-t)}{t}\to 1$$ By L'Hospital's Rule or Taylor series the first fraction tends to $-1$  so that $-1+b-3=1$ or $b=5$ and $a=3$.
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