> 
> 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: 

> 
> $$e^{\frac 45}$$
> 
> 
>