Limit of fibonacci sequence Let $f_n$ be the $n$th Fibonacci number. Find constants $a$ and $b$  such that 
$$\lim_{n\to\infty} \frac{f_n}{a\cdot b^n} = 1$$
I'm somewhat confused on how to approach this problem. I know the closed form of the Fibonnaci sequence, and I think it may have something to do with this problem, but I am unsure of how to proceed. Would love some help!

HINT: You know that
$$f_n=\frac{\varphi^n-\widehat\varphi^n}{\sqrt5}\;,$$
where $\varphi=\frac12\left(1+\sqrt5\right)$ and $\widehat\varphi=\frac12\left(1-\sqrt5\right)$. Note that $|\widehat\varphi|<1$, so $\widehat\varphi^n\to 0$ as $n\to\infty$. Thus, for large $n$ the Fibonacci number $f_n$ is approximately ... ?
