Is answer of limit an exact value?

Solution 1:

$$\lim_{x\rightarrow2}\frac{x^2-4}{x-2}=\lim_{x\rightarrow2}\frac{(x-2)(x+2)}{x-2}=\lim_{x\rightarrow2}(x+2)=4.$$ If $x\rightarrow2$ then $x\neq2$ by the definition of the limit.

Solution 2:

The limit $$\lim_{x\rightarrow 0}\bigg\lfloor \frac{\sin x}{x} \bigg\rfloor = 0 $$

because $\ \bigg\lfloor \displaystyle{\frac{\sin x}{x}} \bigg\rfloor=0\ $ near 0 since the function $\displaystyle{\frac{\sin x}{x}}$ is bounded :

$$ 0 \leq \frac{\sin x}{x} < 1,\ \ \forall x\in[-\pi,\pi] $$

and this implies that:

$$ \ \bigg\lfloor \displaystyle{\frac{\sin x}{x}} \bigg\rfloor=0,\ \ \forall x\neq 0\ \ \text{and}\ \ x\in[-\pi,\pi] $$

and you know that $\ \underset{x\to 0}{\lim}{\displaystyle{\frac{\sin x}{x}}}=1$ and this implies that

$$ \bigg\lfloor\underset{x\to 0}{\lim}{\displaystyle{\frac{\sin x}{x}}}\bigg\rfloor = 1 $$