Can we say $$\lim_{x \to 0} \frac{\sin x}{x} $$ is exactly equal to $1$
I got this doubt while solving the limit
$$\lfloor\lim_{x \to 0}\frac{\sin x}{x}\rfloor$$ because as per my knowledge $\lim_{x \to 0} \frac{\sin x}{x}$ approaches $1$ from below hence
$$\lfloor\lim_{x \to 0}\frac{\sin x}{x}\rfloor=0$$
But some books give it as $1$, so what exactly is the concept here?
I'm assuming you meant $$\lim_{x \to 0}\left\lfloor\frac{\sin x}{x}\right\rfloor=0$$in your question. And indeed, we have both $$\lim_{x \to 0}\left\lfloor\frac{\sin x}{x}\right\rfloor=0\tag{1}$$ and $$\lim_{x \to 0} \frac{\sin x}{x} = 1\tag{2}$$ They are not contradictory! What seems to trouble you is that, for $f(x) = \lfloor x\rfloor$, we have $$\lim_{x \to 0}f\!\left( \frac{\sin x}{x}\right)=0 \neq 1 = f\!\left( \lim_{x \to 0}\frac{\sin x}{x}\right) \tag{3} $$ However, there is no contradiction here. Because $f$ is not continuous at $1$ -- so there is no guarantee that "$f(\lim_n x_n)=\lim_n f(x_n)$" here.