Find the limit:
$$ \lim_{x \rightarrow ( -\frac{1}{10} )^-} [ \frac{1}{x} ]=?$$$[x]$: floor function
my try :
$$\frac{1}{x}-1<\lfloor \frac{1}{x}\rfloor\le \frac{1}{x}$$
$$\lim_{x \rightarrow ( -\frac{1}{10} )^-} \frac{1}{x}-1 =-11$$
$$\lim_{x \rightarrow ( -\frac{1}{10} )^-} \frac{1}{x}-1 =-10$$
!!!!?
Consider a change of coordinates $y = \frac1x $; When $x $ came from the left, $y $ now comes from the right. One way to see it is by thinking of a number to the left of $-\frac1 {10} $, like $-1$. When you flip it with the change of coordinates, $-1 > -10$ so $y $ is going to $-10$ from the right. Hence your limit becomes
$$\lim_{y \to -10^+} \lfloor y \rfloor $$
Notice that as $y $ approaches -10 from the right, that is what $y \to -10^+$ means, it will be always slightly less more than -10. And numbers that are slightly more than -10 have their floor evaluated to -9, as they can be seen as a bit less than $-9$. Thus your limit goes to $-9$. For a better visualization, try plotting the graph $f(x) = \lfloor x \rfloor $ around the point $x=-10$ and imagine yourself walking over the graph, going to -10 from the right. You should see yourself walking in a straight-line at $y=-9$.