Solving $\lim\limits_{x \to 0}\frac{(-1)^{[x]}sin(x^2)}{x}$ 
Solve: $$\lim\limits_{x \to 0}\frac{(-1)^{[x]}\sin(x^2)}{x}$$

Note: $[x]$ is the floor function ( the greatest integer function) . So, usually $(-1)^n$ is bounded between $1$ and $-1$, $[x]$ is bounded $x-1<[x]\le x$ and also $|\sin x^2|\le 1$ . The problem is with $(-1)^{[x]}$ which i am not sure how to evaluate... I am thinking that the answer might be $0$ .
 A: Note you have
$$0 \le \left|\frac{(-1)^{[x]}\sin(x^2)}{x}\right| = \left|x\left(\frac{\sin(x^2)}{x^2}\right)\right| \le |x| \tag{1}\label{eq1A}$$
This uses the hint in Kelenner's question comment that $|\sin(u)| \le |u|$ for all $u \in \mathbb{R}$ (e.g., see Proof that sin(x) ≤ x for All Positive Real Numbers). Thus, you have that as $x \to 0$, the right side of \eqref{eq1A} goes to $0$, so by the squeeze theorem,
$$\lim\limits_{x \to 0}\frac{(-1)^{[x]}\sin(x^2)}{x} = 0 \tag{2}\label{eq2A}$$
A: Since $-1\leq(-1)^{[x]}\leq1$, by squeeze theorem we know either 
$$-\lim_{x\rightarrow0}\frac{\sin x^2}{x}\leq\lim_{x\rightarrow0}\frac{(-1)^{[x]}\sin x^2}{x}\leq\lim_{x\rightarrow0}\frac{\sin x^2}{x}$$ 
or 
$$-\lim_{x\rightarrow0}\frac{\sin x^2}{x}\geq\lim_{x\rightarrow0}\frac{(-1)^{[x]}\sin x^2}{x}\geq\lim_{x\rightarrow0}\frac{\sin x^2}{x}$$ 
(this depends on the sign of $\frac{\sin x^2}{x}$). We can evaluate $\lim_{x\rightarrow0}\frac{\sin x^2}{x}$ by l'Hopital's rule: $$\lim_{x\rightarrow0}\frac{\sin x^2}{x}=\lim_{x\rightarrow0}\frac{2x\cos x^2}1$$
Simplifying further, we have $$\lim_{x\rightarrow0}\frac{2x\cos x^2}1=\lim_{x\rightarrow0}2x\cos x^2=2\cdot0\cdot\cos0^2=2\cdot0\cdot1=0$$ 
Plugging this back in, we have
$$-\lim_{x\rightarrow0}\frac{\sin x^2}{x}\leq\lim_{x\rightarrow0}\frac{(-1)^{[x]}\sin x^2}{x}\leq\lim_{x\rightarrow0}\frac{\sin x^2}{x}\Rightarrow-0\leq\lim_{x\rightarrow0}\frac{(-1)^{[x]}\sin x^2}{x}\leq0$$
And by squeeze theorem, $\lim_{x\rightarrow0}\frac{(-1)^{[x]}\sin x^2}{x}=0$.
A: $$RL=\lim_{x \to 0^+} \frac{(-1)^{[x]} \sin x^2}{x} =\lim_{h \to 0} \frac{(-1)^{[h]} \sin h^2}{h}\frac{h}{h}=\lim_{h \to 0} \frac{(-1)^{[h]} \sin h^2}{h^2} h=0$$
Simolarly,
$$LL+-=L=\lim_{x \to 0^-} \frac{(-1)^{[x]} \sin x^2}{x} =\lim_{h \to 0} \frac{(-1)^{[-h]} \sin h^2}{-h}\frac{h}{h}=\lim_{h \to 0} \frac{(-1)^{[h]} \sin h^2}{h^2} (-h)=0$$
Both the left and the riight limit being finite and equal the required limit exists and it is 0.
