# 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 $\lim\limits_{x\to \infty}\left(\dfrac{P(x)}{5(x-1)}\right)^x$ is?

Attempt:

Using the given conditions I have got, $4a+2b+c=9$ and $6a+b=5$. I am short of an equation to find the function $P(x)$.

For the limit, we have $\left(\dfrac{0}{∞}\right)^∞$ form. Not able to find the way to solve it.

$$e^{\frac 45}$$

$$\lim_{x\to \infty}\left(\dfrac{P(x)}{5(x-1)}\right)^x=\lim_{x\to \infty}\left(\dfrac{ax^2+bx+c}{5x-5}\right)^x=\lim_{x\to \infty}\left(\dfrac{a}{5}x+\dfrac{o(x)}{5(x-1)}\right)^x$$if $a>0$ the limit is $+\infty$ and if $a<0$ the limit doesn't exist therefore $a=0$ and we have$$\lim_{x\to \infty}\left(\dfrac{P(x)}{5(x-1)}\right)^x=\lim_{x\to \infty}\left(\dfrac{bx+c}{5x-5}\right)^x=\lim_{x\to \infty}\left(\dfrac{b}{5}+\dfrac{c+b}{5x-5}\right)^x$$if $b>5$ the limit is $\infty$, for $-5<b<5$ the limit is $0$ and doesn't exist for $b<-5$ therefore $b=5$. Substituting this in the equations you derived yields $c=-1$ and we have$$\lim_{x\to \infty}\left(\dfrac{P(x)}{5(x-1)}\right)^x=\lim_{x\to \infty}\left(1+\dfrac{4}{5x-5}\right)^x=e^{\dfrac{4}{5}}$$

• Your conclusion that $a=0$ is ... bold Commented Jun 16, 2018 at 13:40
• @MostafaAyaz This was really nice Commented Jun 16, 2018 at 13:47
• That's very kind of you! Commented Jun 16, 2018 at 13:48
• I don't follow how you concluded a=0. I mean if a>0 then limit is infinity.. what is the issue with that? Thanks!! Commented Jun 16, 2018 at 14:11
• Sure my friend! I define $f(x)=O(g(x))$ here (the same Big-O notation) if $\lim_{x\to\infty}\dfrac{f(x)}{g(x)}=c$ if $0<c<\infty$. Commented Jun 16, 2018 at 14:56

$P(2)=9$ says $$\tag1 4a+3b+c=9.$$ $P'(3)=5$ says $$\tag26a+b=5$$ $P(\alpha)=0$ says $$\tag3\alpha^2a+\alpha b+c=0$$ Then from $(3)$ and $(1)$, $$\tag4(\alpha^2-4)a+(\alpha-3)b=0$$ so that $|a|\ll |b|$ for $\alpha\gg 0$. Then $(2)$ implies $b\to 5$ and $a \to 0$ as $\alpha\to \infty$, and with $(1)$, $c\to -6$. Hence for all $\alpha\gg 0$, we have $\alpha b+c\sim 5\alpha>0$ and therefore need $a<0$ in $(4)$. Then for such $\alpha$, we have for $x\gg0$ that $\frac{P(x)}{5(x-1)}$ is negative and hence $\left(\frac{P(x)}{5(x-1)}\right)^x$ will in general not make sense

• Equation (1) should read $4a + 2b + c = 9$. Commented Jun 17, 2018 at 7:15

This question is poorly posed, for the reason that the limit to be computed is not made unambiguous: there are two quantities that are changing, $\alpha$ and $x$, and the relationship between their limiting behaviors is not elucidated. For instance, are we meant to compute the iterated limit $$\lim_{x \to \infty} \lim_{\alpha \to \infty} \left(\frac{P(x)}{5(x-1)}\right)^x,$$ or are we to compute $$\lim_{\alpha \to \infty} \lim_{x \to \infty} \left(\frac{P(x)}{5(x-1)}\right)^x,$$ or even $$\lim_{(\alpha,x) \to (\infty,\infty)} \left(\frac{P(x)}{5(x-1)}\right)^x?$$ Or could it be that we are meant to evaluate the limit along a path given by some function $x(\alpha)$ such that as $\alpha \to \infty$, $x(\alpha) \to \infty$? The solution implies the first case, but it is instructive to understand what happens otherwise.

First, we dispense with unnecessary complications. The given restrictions are \begin{align*} 4a + 2b + c &= 9, \\ 6a + b &= 5. \end{align*} This gives us $$b = 5-6a, \quad c = 8a-1,$$ consequently $$a = \frac{1-5\alpha}{8-6\alpha+\alpha^2}.$$ This ultimately gives, in terms of $\alpha$ and $x$, $$\frac{P(x)}{5(x-1)} = \frac{(x-\alpha)(34+\alpha+x-5\alpha x)}{5(x-1)(\alpha-4)(\alpha-2)}.$$ Up to this point, there is nothing preventing us from choosing a relationship between $\alpha$ and $x$: the first iterated limit (the intended problem) is trivial, and the second is clearly infinite. The third therefore does not exist. But we can also see, for example, that if $x = \alpha$, the limit is zero; and if $x = k\alpha$ for $k > 1$, other interesting things happen.