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Le $f$ the fuction: $$f(x)=\left\{\begin{array}{lcl}\frac{\sin(\pi x^2)}{x^2\sin(\pi x)} &\text{if}& x \in \Bbb R \backslash \Bbb Z \\ && \\ \frac 2n &\text{if}&x=n \in \Bbb Z\backslash\{0\} && \end{array}\right.$$

Is the limit $\lim\limits_{x \to +\infty} f(x)$ exists ?

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    Consider how $\frac{\sin( \pi x^2)}{\sin (\pi x)}$ behaves in each of $(n,n+1)$. In particular is it bounded there? – 2017-02-07
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    @user160738: $f$ is bounded in $[n,n+1] $ for all $n \in \Bbb N, n \geq 1$, since $f$ is continuous within the interval and extensible by continuity at the boundaries, but this is not enough to conclude because its extremums depend on $n$ – 2017-02-07

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HINT:

Let $x=n+y$, where $y\in[0,1/2]$. Then,

$$\left|\frac{\sin(\pi x^2)}{\sin(\pi x)}\right|=\left|\frac{\sin\left(2n\pi y\left( 1+\frac{y}{2n}\right) \right)}{\sin(\pi y)}\right|\le \frac{\left|2n\pi y\left(1+\frac{y}{2n}\right) \right|}{\sin(\pi y) }$$

Repeat with $x=(n+1)-z$, where $z\in[0,1/2]$.

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    Thank you Dr.MV. I am inspired by your indication and do the following: 1) Any real $x$ number is written in a unique way: $x=y+n$ where: $n \in \Bbb Z$ and $-1/2 \leq y < 1/2$. 2) For all real number $t$ we have $|\sin(t)|=\sin(|t|)$ 3) For all real numbre $t$ such taht $|t| \leq \frac{\pi}{2}$ we have $\frac{2}{\pi}|t| \leq \sin(|t|) \leq |t|$. – 2017-02-08