$$\ \lim_{x\to 0}\lfloor\tan x\rfloor\cdot \cos x$$
Okay so when I first tried to solve it, I stated at the beginning of the proof
"Note that when
$-\frac{\pi}{4}<x<0, \lfloor\tan(x)\rfloor = -1$
and when
$0<x<\frac{\pi}{4} , \lfloor\tan(x)\rfloor = 0$ "
But I want to learn how to explain it more thoroughly , that will leave no doubt to the grader that I understood the situation. The problem is, I don't quite understand it. I know when we do the one sided limits we can choose a delta such that delta = $\frac{\pi}{4}$, so am I correct in assuming that the expression of the floor can immediately be transformed into $-1/0$ as soon as I choose my delta, and only then I take the limit of the given expression?
To say the limit $\lim_{x\to a} f(x) = L$ is to say for any $\epsilon > 0$ there is a $\delta>0$ so that ....yadda, yadda, yadda.
Now we know that if $\delta_1$ leads to yadda, yadda, yadda then for any $)<\delta < \delta_1$ we also get yadda, yadda, yadda. So we can assume $\lim_{x\to a} f(x)$ that $x$ is as close to $a$ as we need it to be.
And in taking limits we always assume $x \ne a$.
So in calculating $\lim_{x\to 0}f(x)$ we may assume that $-\frac \pi 4 < x < \frac \pi 4$ and that $x\ne a$. (we can also assume $-0.0000000001 < x < 0.087$ and $x \ne 0$... we can assume $x$ is as close to zero as we want. But as long as we assume it is within $\frac \pi 4$ of $0$ we will get something to simplify our work the way we want.)
And that's it
If $-\frac \pi 4 < x < 0$ then $\lfloor\tan x\rfloor\cos x = -\cos x$
And if $0< x < \frac \pi 4$ then $\lfloor\tan x\rfloor\cos x = 0$.
And because we can assume, for the purposes of taking limits that $x$ is within $\frac \pi 4$ of $0$ but not equal to $0$ we have
$\lim_{x\to 0}\lfloor\tan x\rfloor\cos x =$
$\lim_{x\to 0}\begin{cases} -\cos x& x< 0\\ 0&x > 0\end{cases}$
Now if we consider the upper and lower limits:
$\lim_{x\to 0^{-1}}\lfloor\tan x\rfloor\cos x$ we assume $x$ is within $\frac \pi 4$ of $0$; that $x \ne 0$; and we assume $x< 0$.
So
$\lim_{x\to 0^-}\lfloor\tan x\rfloor\cos x =$
$\lim_{x\to 0^-}\begin{cases} -\cos x& x< 0\\ 0&x > 0\end{cases}=$
$\lim_{x\to 0^-}-\cos x = -1$.
And for upper limits we assume $x > 0$ so
$\lim_{x\to 0^+}\lfloor\tan x\rfloor\cos x =$
$\lim_{x\to 0^+}\begin{cases} -\cos x& x< 0\\ 0&x > 0\end{cases}=$
$\lim_{x\to 0^+}0 = 0$.
And as $\lim_{x\to 0^-}\lfloor\tan x\rfloor\cos x \ne \lim_{x\to 0^+} \lfloor\tan x\rfloor\cos x$, $\lim_{x\to 0}[\tan x]\cos x$ doesn't exist.