<s>[INST] If $\lim\limits_{x\to\alpha}\frac{x-2}{x^3-2x+m}=-\infty$, then what are the possible values for $\alpha$ and $m$? 
If $\lim\limits_{x\to\alpha}\dfrac{x-2}{x^3-2x+m}=-\infty$, then what are the possible values for $\alpha$ and $m$?

A student I'm tutoring came to me with this problem. I believe the limit is not one-sided, so that the expression approaches $-\infty$ as $x\to\alpha$ from either direction.
I believe this would require $x=\alpha$ to be a zero of the denominator, so that
$$x^3-2x+m=(x-\alpha)(x^2+\alpha x+\alpha^2-2)+\alpha^3-2\alpha+m$$
The remainder term should vanish, so
$$\alpha^3-2\alpha+m=0$$
But in order for the limit to diverge to the "same" $-\infty$ from either side of $x=\alpha$, I'm under the impression that $x=\alpha$ should actually be a zero of multiplicity $2$. (I'm picturing the behavior of $-\dfrac1{x^2}$ around $x=0$.) Then we'd have
$$x^2+\alpha x+\alpha^2-2=(x-\alpha)(x+2\alpha)+3\alpha^2-2$$
Again the remainder should be $0$, so that
$$3\alpha^2-2=0\implies\alpha=\pm\sqrt{\frac23}\implies m=\pm\frac43\sqrt{\frac23}$$
When I check the limits for either pair of $(\alpha,m)$, I find
$$\lim_{x\to-\sqrt{\frac23}}\frac{x-2}{x^3-2x-\frac43\sqrt{\frac23}}=\color{red}+\infty$$
$$\lim_{x\to\sqrt{\frac23}}\frac{x-2}{x^3-2x+\frac43\sqrt{\frac23}}=-\infty$$
It seems that the sign of $\dfrac{x-2}{x+2\alpha}$ dictates whether the limit diverges to positive or negative infinity.
This explanation seems a bit too hand-wavy and perhaps too verbose for a high-school-level calculus student. Is there a more straightforward or concise argument that can be made to show that $\alpha=\sqrt{\dfrac23}$ and $m=\dfrac43\sqrt{\dfrac23}$ is the answer?
 [/INST] Using the given condition we can see that $(x^3-2x+m)/(x-2)$ is negative and tends to $0$ as $x\to a$ (replaced $\alpha$ with $a$ to simplify typing). Note that the given condition also excludes $a=\pm\infty$ and hence we assume $a\in\mathbb {R} $. Then by multiplication with $(x-2)$ we get $$\lim_{x\to a} (x^3-2x+m)=0$$ so that $$a^3-2a+m=0\tag{1}$$ If $a=2$ then $m=-4$ and then $$x^3-2x-4=(x-2)(x^2+2x+2)$$ and then $(x^3-2x+m)/(x-2)$ does not tend to $0$. Hence $a\neq 2$.
From $(1)$ we can see that $$x^3-2x+m=x^3-2x-a^3+2a=(x-a)(x^2+ax+a^2-2)$$ and we need to ensure that the expression above maintains a constant sign opposite to that of $(a-2)$ as $x\to a$. This is possible only when the factor $x^2+ax+a^2-2$ has a single root $a$ ie $3a^2-2=0$ or $a=\pm\sqrt{2/3}$ and $$x^2+ax+a^2-2=(x-a)(x+2a)$$ so that $x=a$ is indeed a single root.
Next note that we have $$x^3-2x+m=(x-a)^2(x+2a)$$ and we want its sign to be opposite to that of $a-2$ as $x\to a$ so that $a=\sqrt{2/3}$ is the only option. 
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