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?
We claim that the limit as $x \to 2-$ is $1$; but:

If $0 &lt; 2-x &lt; 1/2$, then
$$
|\{ 2x - \lfloor x \rfloor^{2} \} - 1| = \lfloor 2x-1 \rfloor - 2x+2 = 4-2x = 2(2-x);
$$
given any $\varepsilon &gt; 0$, we have $2(2-x) &lt; \varepsilon$ if in addition $2-x &lt; \varepsilon/2$. Hence, for every $\varepsilon &gt; 0$, if $0 &lt; 2-x &lt; \min \{1/2, \varepsilon/2 \}$ then $|\{ 2x - \lfloor x \rfloor^{2} \} - 1| &lt; \varepsilon$.
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.