Dtyjlui

If $limlimits_{xtoalpha}dfrac{x-2}{x^3-2x+m}=-infty$, 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 $xtoalpha$ from either direction.

I believe this would require $x=alpha$ to be a zero of the denominator, so that
$$x^3-2x+m=(x-alpha)(x^2+alpha x+alpha^2-2)+alpha^3-2alpha+m$$
The remainder term should vanish, so
$$alpha^3-2alpha+m=0$$
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
$$x^2+alpha x+alpha^2-2=(x-alpha)(x+2alpha)+3alpha^2-2$$
Again the remainder should be $0$, so that
$$3alpha^2-2=0impliesalpha=pmsqrt{frac23}implies m=pmfrac43sqrt{frac23}$$
When I check the limits for either pair of $(alpha,m)$, I find
$$lim_{xto-sqrt{frac23}}frac{x-2}{x^3-2x-frac43sqrt{frac23}}=color{red}+infty$$
$$lim_{xtosqrt{frac23}}frac{x-2}{x^3-2x+frac43sqrt{frac23}}=-infty$$
It seems that the sign of $dfrac{x-2}{x+2alpha}$ 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 $m=dfrac43sqrt{dfrac23}$ is the answer?

Your proof is good, but it can be simplified using some other facts.

We cannot have $alpha=pminfty$, because the limit there is $0$.

Thus $alpha$ must be a root of the denominator and actually of multiplicity $2$, because otherwise the limit would be $infty$ from one side and $-infty$ from the other side. Unless the root is $2$ and has multiplicity $3$; this is not possible, because $2$ is a root only if $m=-4$ and the denominator has three distinct roots.

How can $alpha$ be a root of multiplicity $2$ (or $3$)? It must also be a root of the derivative $3x^2-2$. Hence it must be either $sqrt{2/3}$ or $-sqrt{2/3}$.

Why is this true? Suppose $p(x)=(x-alpha)^2q(x)$ ($p$ any polynomial). Then $p'(x)=2(x-alpha)q(x)+(x-alpha)^2q'(x)$, so $p'(alpha)=0$. Conversely, suppose that $p(alpha)=p'(alpha)=0$; then $p(x)=(x-alpha)^2q(x)+ax+b$ (long division). The condition $p(alpha)=0$ implies $aalpha+b=0$; the condition $p'(alpha)=0$ implies $a=0$. Thus also $b=0$ and $alpha$ is a root of $p$ with multiplicity at least $2$.

The case $alpha=sqrt{2/3}$ gives
$$
m=2alpha-alpha^3=sqrt{frac{2}{3}}left(2-frac{2}{3}right)=frac{4}{3}sqrt{frac{2}{3}}
$$
The case $alpha=-sqrt{2/3}$ gives
$$
m=2alpha-alpha^3=-sqrt{frac{2}{3}}left(2-frac{2}{3}right)=-frac{4}{3}sqrt{frac{2}{3}}
$$
What's the other root? The sum of the roots is $0$ (Viète’s formulas), so it's $-2alpha$. We need that
$$
frac{alpha-2}{alpha-2alpha}<0
$$
so that the limit is $-infty$. This means
$$
frac{2}{alpha}-1<0
$$
Clearly $alpha<0$ satisfies the requirement. If $alpha>0$, we must have $alpha>2$. However the positive value for $alpha$ has
$$
alpha^2=frac{32}{27}<4
$$
so it is not valid.

No hand-waving. If $(alpha-2)/(-alpha)<0$ the given function is negative in a whole punctured neighborhood of $alpha$; it has infinite limit because the denominator vanishes, so the limit is $-infty$.

Using the given condition we can see that $(x^3-2x+m)/(x-2)$ is negative and tends to $0$ as $xto a$ (replaced $alpha$ with $a$ to simplify typing). Note that the given condition also excludes $a=pminfty$ and hence we assume $ainmathbb {R} $. Then by multiplication with $(x-2)$ we get $$lim_{xto a} (x^3-2x+m)=0$$ so that $$a^3-2a+m=0tag{1}$$ 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 $aneq 2$.

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 $xto a$. This is possible only when the factor $x^2+ax+a^2-2$ has a single root $a$ ie $3a^2-2=0$ or $a=pmsqrt{2/3}$ 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 $xto a$ so that $a=sqrt{2/3}$ is the only option.