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 ?
 A: Hint.
Note that $\lfloor z \rfloor=0$ for $0\leqslant z<1$. On the other hand,
$$
\frac{1}{1+x}\in[0,1)
$$
for $x>0$ and in particular for $x$ near $1$.
A: $$\lim_{x \to 1^-} (x+1) \left\lfloor \frac{1}{x+1}\right\rfloor $$
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
A: 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 
$$ \color{red}{\lim_{x \to 1^-}(x+1)\left\lfloor \frac{1}{x+1}\right\rfloor  =0}~$$
A: There is no real difficulty here. The first factor tends to $2$, and the fraction inside the floor to $\dfrac12$. Hence 
$$\lim_{1^-}=\lim_{1^+}=0.$$
