The priority of limits How are the two expressions different?
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}}{x}\bigg\rfloor$$
and $$\bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor$$
If limit is inside the floor function, Do I have to apply the limits first?
If this is the case, then, $$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}}{x}\bigg\rfloor=0$$
$$\bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor=1$$
Am I solving this right?
Also how can I calculate,
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor$$ 
Thank you.
 A: $$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}}{x}\bigg\rfloor$$
In this first, you have to take the floor of the function then apply the limit on its floor.
$$\frac{\sin{x}}{x}< 1$$ when $x \to 0$
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}}{x}\bigg\rfloor=0$$
$$\bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor$$
Here you have to first calculate the limit then take the floor of it. You know that this limit is 1 thus floor of 1 is 1.
$$\bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor=1$$
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor$$
Series expansion at $x=0$ gives 
$$1+\frac{x^2}{6}+\frac{31x^4}{360}+O(x^6)$$
$$\frac{\sin{x}\cdot \tan{x}}{x^2}\ge 1$$
$$\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor=1$$
thus 
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor=1$$
A: The first expression says:


*

*Take the function $f(x) = \lfloor\frac{\sin(x)}{x}\rfloor$.

*Take the limit $$\lim_{x\to 0} f(x)$$


The second expression says:


*

*Take the function $f(x) = \frac{\sin(x)}{x}$.

*Take the limit $$L=\lim_{x\to 0} f(x)$$.

*Calculate $\lfloor L\rfloor$


The two expressions do not need to be the same.
A: Yes you are right indeed since
$$\frac{\sin x}{x}<1 \implies \bigg \lfloor\frac{\sin{x}}{x}\bigg\rfloor=0 \implies \lim_{x\to0} \bigg \lfloor\frac{\sin{x}}{x}\bigg\rfloor=0$$
and since
$$\lim_{x\to0} \frac{\sin{x}}{x}=1 \implies \bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor=1
$$
For the latter for $x$ sufficiently small we have 
$$1<\frac{\sin{x}\cdot \tan{x}}{x^2}<2$$
indeed $\sin x>x-\frac{x^3} 6$ and $\tan x>x+\frac{x^3} 3$ and 
$$\frac{\sin{x}\cdot \tan{x}}{x^2}>1+\frac{x^2}6-\frac{x^6}{18}$$
therefore
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor=1$$
A: Here's fact 4
If $f(x)$ is continuous at $a$ and $\mathop {\lim }\limits_{x \to a} g\left( x \right) = b$ then:
$\mathop {\lim }\limits_{x \to a} f\left( {g\left( x \right)} \right) = f\left( {\mathop {\lim }\limits_{x \to a} g\left( x \right)} \right)$
Floor function isn't continuous, so you should apply floor function first, before calculating limit. 
Let's calculate 
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor$$
Okay, the first thing I want to do it's to solve:
$$\lim_{x\to0}\frac{\sin{x}\cdot \tan{x}}{x^2} = \lim_{x\to0} \frac{\sin{x}}{x}{\frac{\sin{x}}{x}}{\frac{1}{\cos{x}}} = [\lim_{x\to0}\frac{\sin{x}}{x} = 1] = \lim_{x\to0}\frac{1}{\cos{x}} = 1$$
So, 
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor = 
\lim_{x\to0}\bigg\lfloor\frac{1}{\cos{x}}\bigg\rfloor 
$$
$$\cos{x} \le 1 $$$$\frac{1}{\cos{x}} \ge 1$$
$$x < \arccos(2) => \cos{x} < \frac{1}{2} => \frac{1}{\cos{x}}< 2$$
It means that in the neighborhood of 0 with radius $\delta < arccos(2)$ function $\bigg\lfloor\frac{1}{\cos{x}}\bigg\rfloor \equiv 1$ 
And we can conclude that 
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor =
\lim_{x\to0}\bigg\lfloor\frac{1}{\cos{x}}\bigg\rfloor 
= 1
$$
