Limit with fractional part and greatest integer part 
Find $$\lim_{x\rightarrow 2-} \{x+(x-[x]^2)\}$$

For $x\to 2-$, $[x]=1$, i.e $[x]^2=1$, 
so $\lim_{x\rightarrow 2-} \{x+(x-1)\}=\lim_{x\rightarrow 2-} \{2x-1\}=3.$
I don't know whether $\{\}$ symbolizes fractional part=$x-[x]$ or not for this particular question. I have assumed  $\{\}$ not as fractional part and solved. Please check my solution is correct or not.

In case $\{\}$ symbolizes fractional part=$x-[x]$, then what will be the answer? 

 A: We claim that the limit as $x \to 2-$ is $1$; but:
If $0 < 2-x < 1/2$, then
$$
|\{ 2x - \lfloor x \rfloor^{2} \} - 1| = \lfloor 2x-1 \rfloor - 2x+2 = 4-2x = 2(2-x);
$$
given any $\varepsilon > 0$, we have $2(2-x) < \varepsilon$ if in addition $2-x < \varepsilon/2$. Hence, for every $\varepsilon > 0$, if $0 < 2-x < \min \{1/2, \varepsilon/2 \}$ then $|\{ 2x - \lfloor x \rfloor^{2} \} - 1| < \varepsilon$.
A: Let's assume that $\{\} $ denotes fractional part so that $\{x\} =x-[x] $. And then we can see that as $x\to 2^{-}$ we have $[x] =1$ and hence the expression under limit is $\{2x-1\} $ which is same as $2x-1-[2x-1]$. And as $x\to 2^{-}$ we have $[2x-1]=2$ so that the expression under limit is $2x-3$ which tends to $1$ as $x\to 2^{-}$.
A: If {} denotes curl brackets, then your answer is correct.
However if {} mean fractional part function, then the limit will tend to $0$ as if $x\to 2^{-}$, then $2x\to 4^{-}$ and $2x-1\to 3^{-}$ and $\{2x-1\}\to \{3^{-}\}$.
Now $0 \le \{3^{-}\} < 1$ . 
As limiting case, you can however consider $\{3^{-}\}=0$ when $x=2$.
A: Fixing $1<x<2$ we always have $[x] = 1$ hence 
$$x+(x -[x]^2) = 2x -1~~~~\forall~~ x\in (1,2)$$
therefore, 
$$\lim_{x\rightarrow 2-} \{x+(x-[x]^2)\} = \lim_{x\rightarrow 2-}2x-1 = 3$$
A: lim −{x+(x−[x]2)}=1, Because {2x-1} lies between 0 to 1, but in 
x→2                  question 
                     it was mentioned as 2 minus, hence it will
                     lead to 1.
