Consider the Limit
$$ L_1 = \lim_{x \to 0}\left\lfloor\frac{\sin x}{x}\right\rfloor . $$
We have $$ L_1 = \lim_{x \to 0} \left\lfloor \frac{x-\frac{x^3}{6} + \dots}{x} \right\rfloor = \lim_{x \to 0}\left\lfloor{1-\frac{x^2}{6}+\cdots}\right\rfloor . $$ Now my Doubt is what value we will take for $1-\frac{x^2}{6}$ because as per my knowledge limit cannot be taken inside the Greatest integer function.
But where as the Limit
$$L_2 = \left\lfloor\lim_{x \to 0}\frac{\sin x}{x}\right\rfloor=0$$
So are the two limits essentially same, if same does not it mean limit and Greatest integer function are interchanged?
You have $[1]=1$. Hence $L_2=1$ as $\lim\limits_{x \to 0} \frac{\sin x}{x}=1$.
However $L_1 =0$ as for $x\neq 0$ you have
$$0 < \frac{\sin x}{x} <1.$$
$L_1 \neq L_2$ here and you can’t exchange $\lim$ and $[\cdot ]$ in that case.