# Flaw in evaluating $\lim_{x\to \infty}\left(\frac{P(x)}{5(x-1)}\right)^x$

There was this question asked yesterday here: Link: Evaluating $\lim\limits_{x→∞}\left(\frac{P(x)}{5(x-1)}\right)^x$

Consider $P(x)= ax^2+bx+c$ where $a,b,c \in \mathbb R$ and $P(2)=9$. Let $\alpha$ and $\beta$ be the roots of the equation $P(x)=0$.

If $\alpha \to \infty$ and $P'(3)= 5$, then $\lim_{x\to \infty}\left(\dfrac{P(x)}{5(x-1)}\right)^x$ is ?

$e^{\frac 45}$

I gave a thought to it and this is what I did:

If $\alpha \to \infty$ and $\alpha$ is a root of $P(x)$, then for $x \to \infty$, $P(x) \to 0$.

Also, as $x \to \infty$, the denominator $5(x-1) \to \infty$.

Therefore, $\dfrac {P(x)}{5(x-1)} \to 0$. And hence the whole limit asked in the question tends to zero.

There is a flaw in my reasoning. I am trying to find it. Is that flaw that solving this limit part by part is illegal? Or, is it something else? I am still learning it, any help would be really appreciated. Thank you.

• Why is the first sentence true? (If $\alpha \to \infty$ and ...)
– sku
Jun 17, 2018 at 5:46
• The condition $\alpha \to \infty$ is really unclear to me.
– user
Jun 17, 2018 at 5:58
• @sku I said that because the question mentions that $\alpha$ is a root of the equation and it tends to infinity. Jun 17, 2018 at 5:59
• @gimusi Yes, I am confused too. I think you might find more on the link I attached, the original question. They say that if one root of quadratic is infinity, the equation becomes linear. Jun 17, 2018 at 6:00
• @IceInkberry I really doubt that the equation becomes linear if one root tends to infinity. Jun 17, 2018 at 6:09

The problem with your reasoning is rooted, so to speak, in the fact that the problem, as stated both here and in the original link, is very poorly posed. It barely makes sense to talk about the limit of a polynomial as one of its roots tends to infinity, and whatever that notion means, it's absent from the expression $\lim_{x\to\infty}\left(P(x)\over5(x-1)\right)^x$. I suspect what's really meant is the following double limit:

$$\lim_{x\to\infty}\left(\lim_{\alpha\to\infty}(a(\alpha)x^2+b(\alpha)x+c(\alpha))\over5(x-1) \right)^x$$

where $a=a(\alpha),b=b(\alpha),c=c(\alpha)$ are (the unique) solutions of the system of equations

\begin{align} \alpha^2a+\alpha b+c&=0\qquad\text{i.e., }P(\alpha)=0\\ 4a+2b+c&=9\qquad\text{i.e., }P(2)=9\\ 6a+b&=5\qquad\text{i.e., }P'(3)=5 \end{align}

In other words, before you take the limit as $x$ goes to infinity, you have to solve for the coefficients of $P$ as functions of $\alpha$, then take the limit of what you get as $\alpha$ goes to infinity, and only then take the limit of what's left as an expression in the variable $x$. The answer by heropup at the original, linked question, discusses all this.

As for what's wrong in your reasoning, it has a certain superficial logic to it. Indeed, when writing a limit expression, you can use any variable you like, so $\lim_{x\to\infty}P(x)=\lim_{\alpha\to\infty}P(\alpha)$, at which point one is tempted to say that, since $\alpha$ is a root of $P$, which we want to take to infinity, we have $$\lim_{x\to\infty}P(x)=\lim_{\alpha\to\infty}P(\alpha)=\lim_{\alpha\to\infty}0=0$$

The trick is, when picking a dummy variable to express a limit (or a sum, or whatever), you're not supposed to pick a variable that's already been assigned a meaning. So when I said you can use any variable you like, that was a lie; there are rules that limit, so to speak, how you can express the taking of a limit. Usually the restrictions are obvious; in this case they're not, but that's mostly because of how poorly posed the problem is.