Question

What is alpha and m

Answer 1

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
`x=\alpha` be a root of the denominator. But more than that, because the limit appears to be -infinity as
`x\to\alpha` from either side, the root has to have even multiplicity.

Polynomial division shows that

`x^3-2x+m=(x-\alpha)(x^2+\alpha x+\alpha^2-2)+\alpha^3-2\alpha+m`

Since
`x=\alpha` is a root, the remainder term will vanish, so we know

`\alpha^3-2\alpha+m=0`

Another round of division on the quotient term above shows that

`x^2+\alpha x+\alpha^2-2=(x-\alpha)(x+2\alpha)+3\alpha^2-2`

Again, the remainder term will vanish, so

`3\alpha^2-2=0\implies\alpha=\pm√(\frac23)`

which in turn forces

`\alpha^3-2\alpha+m=0\implies m=\pm\frac43√(\frac23)`

So we've rewritten the limit as

`\displaystyle\lim_(x\to\alpha)(x-2)/(x^3-2x+m)=\lim_(x\to\alpha)(x-2)/((x-\alpha)^2(x+2\alpha))`

However, notice that if
`\alpha=-√(\frac23)`, we have

`(x-2)/(x+2\alpha)>0`

and it's this expression's sign that would force the
`\frac1{(x-\alpha)^2}` part of the limand to diverge to *positive* infinity. This doesn't happen if we take the other choice of
`\alpha=√(\frac23)`, since

`(x-2)/(x+2\alpha)<0`

forces a negative sign as
`\frac1{(x-\alpha)^2}` diverges to infinity, so the overall limit is *negative* infinity.