<s>[INST] How to find $\lim_{x \to 1^-} (x+1) \lfloor \frac{1}{x+1}\rfloor $? find the limits :
$$\lim_{x \to 1^-} (x+1) \lfloor \frac{1}{x+1}\rfloor =?$$

My try :
$$\lfloor \frac{1}{x+1}\rfloor=\frac{1}{x+1}-p_x \ \ \ : 0\leq p_x <1$$
So we have :
$$\lim_{x \to 1^-} (x+1) (\frac{1}{x+1}-p_x) \\= \lim_{x \to 1^-} -(x+1)p_x=!??$$
Now what ?
 [/INST] For $x$ near $1$ let say $$\frac12<x<1 \Longleftrightarrow  -\frac12<x-1<0$$ we have have that 
$$ \frac12<x<1 \Longleftrightarrow \frac32<x+1<2  \Longleftrightarrow \frac12 <\frac{1}{x+1} <\frac 23$$
Therefore 
$$ \left\lfloor \frac{1}{x+1}\right\rfloor  =0~~~\forall~ \frac12<x<1 $$
That is , 
$$ (x+1)\left\lfloor \frac{1}{x+1}\right\rfloor  =0~~~\forall~ \frac12<x<1 $$
that is 
$$ \color{red}{\lim_{x \to 1^-}(x+1)\left\lfloor \frac{1}{x+1}\right\rfloor  =0}~$$
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