find the limits $\lim_{x\to \infty} \lfloor f(x) \rfloor=? $

Question

let $f(x)=\dfrac{4x\sin^2x-\sin 2x}{x^2-x\sin 2x}$ the fine the limits :

$$\lim_{x\to \infty} \lfloor f(x) \rfloor=? $$

$$f(x)=\dfrac{4x\sin^2x-\sin 2x}{x^2-x\sin 2x}=\dfrac{2\sin x(2x\sin x-\cos x)}{x(x-\sin2x)}$$

what do i do ?

Answer 1

A more heuristic argument is as follows: The key observation is that $\sin u$ is always between $-1$ and $1$, no matter what $u$ is. Consider the impact of this on the numerator. We have

\begin{align} \text{numerator} & = 4x \times \text{(something between $-1$ and $1$)} - \text{(something between $-1$ and $1$)} \\ & = \text{(something between $-4|x|$ and $4|x|$)} - \text{(something between $-1$ and $1$)} \\ & = \text{(something between $-4|x|-1$ and $4|x|+1$)} \end{align}

Now, consider the impact of that observation on the denominator. We have

\begin{align} \text{denominator} & = x^2 - x \times \text{(something between $-1$ and $1$)} \\ & = x^2 - \text{(something between $-|x|$ and $|x|$)} \\ & = \text{(something between $x^2-|x|$ and $x^2+|x|$)} \end{align}

So our fraction consists of something which is linear in $x$, divided by something that is quadratic in $x$. Therefore, the limit must be $0$, as $x$ increases without bound (in either direction).

Answer 2

Let $g(x) = 4x\sin(x)^2 - \sin(2x)$ be the numerator of $f(x)$. For any $n \in \mathbb{Z}_{+}$, let $\epsilon_n$ be a small positive number such that $$0 < \epsilon_n < \min\left(\frac{1}{\pi(n\pi+1)}, \frac{\pi}{4}\right)$$

Consider the sequence $x_n = n\pi + \epsilon_n$ which diverges to infinity, we have $$g(x_n) = 4(n\pi + \epsilon_n)\sin(n\pi + \epsilon_n)^2 - \sin(2n \pi + 2\epsilon_n) \le 4(n\pi + 1)\sin(\epsilon_n)^2 - \sin(2\epsilon_n)$$

Notice for $y \in (0,\frac{\pi}{2})$, we have $\frac{2}{\pi}y < \sin(y) < y$. We can bound above value as $$g(x_n) \le 4(n\pi+1)\epsilon_n^2 - \frac{4}{\pi}\epsilon_n = 4(n\pi+1)\epsilon_n\left(\epsilon_n - \frac{1}{\pi(n\pi + 1)}\right) < 0$$

Since the denominator $x^2 - x\sin(2x)$ is positive whenever $x > 1$, we find

$$f(x_n) < 0 \quad\implies\quad \lfloor f(x_n) \rfloor \le -1$$

If $\ell = \lim_{x\to\infty} \lfloor f(x)\rfloor$ exists, this will lead to $$\ell = \lim_{n\to\infty} \lfloor f(x_n)\rfloor \le -1$$

On the other hand, $f(n\pi) = 0$ for all $n \in \mathbb{Z}_{+}$. If $\ell$ exists, we will have

$$\ell = \lim_{n\to\infty}\lfloor f(n\pi)\rfloor =\lim_{n\to\infty} 0 = 0$$

This implies $0 \le -1$ which is nonsense! As a result, $\lim_{n\to\infty}\lfloor f(x)\rfloor$ cannot exist!