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

Question

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?

Answer 1

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

We cannot have $\alpha=\pm\infty$, 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 $a\alpha+b=0$; 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 $$ m=2\alpha-\alpha^3=\sqrt{\frac{2}{3}}\left(2-\frac{2}{3}\right)=\frac{4}{3}\sqrt{\frac{2}{3}} $$ The case $\alpha=-\sqrt{2/3}$ gives $$ m=2\alpha-\alpha^3=-\sqrt{\frac{2}{3}}\left(2-\frac{2}{3}\right)=-\frac{4}{3}\sqrt{\frac{2}{3}} $$ What's the other root? The sum of the roots is $0$ (Viète’s formulas), so it's $-2\alpha$. We need that $$ \frac{\alpha-2}{\alpha-2\alpha}<0 $$ 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 $-\infty$.

Answer 2

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.