Is the function $f(x)=\frac{\sin x}{x}$ with domain $D_f=\mathbb{R}\setminus\{0\}$, monoton or convex from a number on? Is the function $f(x)=\frac{\sin x}{x}$ with domain $D_f=\mathbb{R}\setminus\{0\}$, monoton or convex from a number on?
I know that $f$ is non-negative and decreasing  over $\left[0,\pi\right]$, but does there exist a real number $M$ such that $f$ can be monoton or convex on $(M,+\infty)$?  
 A: No. It will keep oscillating up and down with a period of $2\pi$, although with smaller and smaller amplitude (so it's not truly periodic).
One way to see this is to just look for zeroes:
$$
\begin{align}
\frac{\sin x}x &= 0\\
\frac{\sin x}{x}\cdot x& = 0\cdot x,\quad \text{assuming that }x\neq 0\\
\sin x &= 0
\end{align}
$$
so for any $x$ other than $0$, the two functions $\sin x$ and $\frac{\sin x}{x}$ are zero in exactly the same places, which is to say on any integer multiple of $\pi$ and nowhere else. Those integer multiples never stop no matter how far out on the number line you go, so the function never ceases its oscillation.
A: For $x > 0$, the derivative of $f(x)$ is $f^{\prime}(x) = \tfrac{\cos x}{x} - \tfrac{\sin x}{x^{2}}$. Since $\cos x$ and $\sin x$ are bounded in norm by $1$, for large $x$ the term $\tfrac{\cos x}{x}$ dominates and the term $-\tfrac{\sin x}{x^{2}}$ is negligible. So one expects the derivative to change sign infinitely often as $x \to \infty$. More concretely, $f^{\prime}(2k\pi) = \tfrac{1}{2k\pi}$ and $f^{\prime}((2k+1)\pi) = -\tfrac{1}{(2k+1)\pi}$ for $k$ a positive integer. 
