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. 
The answer is: 

$$e^{\frac 45}$$

 A: $$\lim_{x\to
 \infty}\left(\dfrac{P(x)}{5(x-1)}\right)^x=\lim_{x\to
 \infty}\left(\dfrac{ax^2+bx+c}{5x-5}\right)^x=\lim_{x\to
 \infty}\left(\dfrac{a}{5}x+\dfrac{o(x)}{5(x-1)}\right)^x$$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$$\lim_{x\to
 \infty}\left(\dfrac{P(x)}{5(x-1)}\right)^x=\lim_{x\to
 \infty}\left(1+\dfrac{4}{5x-5}\right)^x=e^{\dfrac{4}{5}}$$
A: $P(2)=9$ 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
A: 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.  
