According to Wolfram Alpha : $\lim_{x \to 0^{-}}\lfloor x \rfloor = -1$ and $\lfloor\lim_{x \to 0^{-}} x\rfloor = 0$ . The first expression is obvious but the second doesn't make sense . It should be $-1$ because for example we have $\lfloor - 0.00001 \rfloor = -1$ . My teacher also accepted the Wolfram Alpha's result . I'm really confused about it .
2 Answers
Hint: $\lim_{x\rightarrow 0^{-}}{x}=0$.
You first evaluate the limit and then take the floor value.
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1$\begingroup$ How is this a hint?? This post is pretty much a full solution XD $\endgroup$ Aug 10, 2017 at 22:43
This is due the fact that the function $x \mapsto \lfloor x\rfloor$ isn't conutinuous. You have to be aware of how a limit works. For the first term $\lim_{x\to 0-} \lfloor x\rfloor$ you take values which have a small distance to $0$ and are negative. In other words for every $\epsilon>0$ you take $y \in (-\epsilon,0)$ and for all these $y$ you have $$ \lfloor y \rfloor = -1. $$ Hence, no matter how close you come to $0$ the value of the bracket will always be $-1$, which means nothing else than the limit is also $-1$.
On the other hand we have $\lim_{x\to 0 -} x= 0$ which is pretty obvious. Now if you insert that into the floor brackets, you obatin $$\lfloor\lim_{x\to 0 -} x\rfloor = \lfloor0\rfloor=0$$