<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] $$\lim_{x \to 1^-} (x+1) \left\lfloor \frac{1}{x+1}\right\rfloor $$
Let ${x = 1-y}$ 
So our equation is same as
$$\lim_{y \to 0^+} (2-y) \left\lfloor \frac{1}{2-y}\right\rfloor $$
Floor value is $0$. Also value outside floor funtion tends to $2$.
Thus our equation becomes 
$2\cdot 0=0$
So answer is $0$.
QED
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