Let $P(x)$ be any polynomial with positive real coefficients. Determine the following limit with proof
$ \lim \limits_{x \to \infty} \dfrac{\lfloor P(x) \rfloor}{P(\lfloor x \rfloor)}$
I am unable to approach these kind of sums. Anyone please can guide me and help me with this solution?
Here, we assume $P(x)$ is nonconstant. Your limit is equal to, for some $a,b\in[0, 1)$ ($a$ and $b$ are also functions of $x$):
$$\lim_{x\to\infty}\frac{P(x) - a}{P(x - b)}$$
Note that we can rearrange this as:
$$\lim_{x\to\infty}\frac{P(x) - a}{P(x)}\frac{P(x)}{P(x-b)} = \lim_{x\to\infty}\frac{P(x) - a}{P(x)}\lim_{x\to\infty}\frac{P(x)}{P(x - b)}$$
$$\lim_{x\to\infty}\bigg(1 - \frac{a}{P(x)}\bigg)\lim_{x\to\infty}\frac{P(x)}{P(x-b)}$$
Then, since $P(x)$ has positive coefficients and is nonconstant, $P(x)\to\infty$ as $x\to\infty$. Thus, since $a$ is bounded:
$$\lim_{x\to\infty}\bigg(1 - \frac{a}{P(x)}\bigg)\lim_{x\to\infty}\frac{P(x)}{P(x-b)} = \bigg(1 - \frac{a}{\infty}\bigg)\lim_{x\to\infty}\frac{P(x)}{P(x-b)} = \lim_{x\to\infty}\frac{P(x)}{P(x-b)}$$
Then, taking $x$ at the integers, which are also unbounded in the positive direction, we find that $b = 0$, so:
$$\lim_{x\to\infty}\frac{P(x)}{P(x-b)}=\lim_{x\to\infty, x\in\mathbb{Z}}\frac{P(x)}{P(x-b)} = \lim_{x\to\infty, x\in\mathbb{Z}}\frac{P(x)}{P(x)} = 1$$ Thus:
$$\boxed{\lim_{x\to\infty}\frac{\lfloor P(x)\rfloor}{P(\lfloor x\rfloor)} = 1}$$
In the case of the constant function $P(x) = p$, then $\lim_{x\to\infty}\frac{\lfloor P(x)\rfloor}{P(\lfloor x\rfloor)} = \frac{\lfloor p\rfloor}{p}$, which is not always equal to $1$.