Since the degree of the numerator (1) is smaller than the degree of the denominator (3), the only way for the limit to diverge to -infinity is to have [tex]x=\alpha[/tex] be a root of the denominator. But more than that, because the limit appears to be -infinity as [tex]x\to\alpha[/tex] from either side, the root has to have even multiplicity.
Polynomial division shows that
[tex]x^3-2x+m=(x-\alpha)(x^2+\alpha x+\alpha^2-2)+\alpha^3-2\alpha+m[/tex]
Since [tex]x=\alpha[/tex] is a root, the remainder term will vanish, so we know
[tex]\alpha^3-2\alpha+m=0[/tex]
Another round of division on the quotient term above shows that
[tex]x^2+\alpha x+\alpha^2-2=(x-\alpha)(x+2\alpha)+3\alpha^2-2[/tex]
Again, the remainder term will vanish, so
[tex]3\alpha^2-2=0\implies\alpha=\pm\sqrt{\dfrac23}[/tex]
which in turn forces
[tex]\alpha^3-2\alpha+m=0\implies m=\pm\dfrac43\sqrt{\dfrac23}[/tex]
So we've rewritten the limit as
[tex]\displaystyle\lim_{x\to\alpha}\frac{x-2}{x^3-2x+m}=\lim_{x\to\alpha}\frac{x-2}{(x-\alpha)^2(x+2\alpha)}[/tex]
However, notice that if [tex]\alpha=-\sqrt{\dfrac23}[/tex], we have
[tex]\dfrac{x-2}{x+2\alpha}>0[/tex]
and it's this expression's sign that would force the [tex]\dfrac1{(x-\alpha)^2}[/tex] part of the limand to diverge to *positive* infinity. This doesn't happen if we take the other choice of [tex]\alpha=\sqrt{\dfrac23}[/tex], since
[tex]\dfrac{x-2}{x+2\alpha}<0[/tex]
forces a negative sign as [tex]\dfrac1{(x-\alpha)^2}[/tex] diverges to infinity, so the overall limit is *negative* infinity.
