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

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).
