Let:
$$f(x) := \begin{cases} x+3 &\text{, if } x\in (-2,0) \\ 4 &\text{, if } x=0 \\ 2x+5 &\text{, if } 0<x<1\end{cases}\;.$$
Then find $\lim_{x\to 0^-} f([x-\tan x])$, where $[\cdot]$ is greatest integer.
Since $x$ is approaching zero from the negative side, $x>\tan x$
Therefore $[x-\tan x] =0$
$$\lim_{x\to 0^-}f( [x-\tan x])=[x+\tan x] +3 =3$$
But the right answer is 4.
Basically, I am having a problem in choosing the right function for the given limit, because $f(x)=4$ for $x=0$, but $x$ actually isn’t 0, then how can it be the answer?
Best Answer
For $-\pi/2 < x < 0$ you have $\tan x < x < 0$, hence $x - \tan x > 0$. On the other hand, in a suitable left neighborhood $-\sigma < x <0$ of $0$ (with $\sigma > 0$), you have also $2x < \tan x$, therefore $x - \tan x < -2x$; moreover you can also choose $\sigma$ so small that $-2x < 1$. Thus $-\sigma < x < 0 \Rightarrow 0 < x - \tan x < 1$ and this entails $[x - \tan x] = 0$ everywhere in the small left neighborhood $]-\sigma , 0[$ of $0$.
Finally, you find:
$$\lim_{x \to 0^-} f([x - \tan x]) = \lim_{x\to 0^-} f(0) = f(0)\; ,$$
therefore you have to check the text of your exercise: if $f(0) = 0$ as you wrote at the beginning, the answer is $0$; if $f(0) = 4$ as you wrote at the end, the answer is $4$. ;-)