where $[x]$ denotes step up function $\& \{x\}$ fractional part function.
Let's check right hand limit at $c=3$
$x=3+h$
$\lim _{h \rightarrow 0} f(3+h)=\lim _{h \rightarrow 0}[[3+h]]-[2(3+h)-1]=4-6=-2$
Let's check Left hand limit at $c=3$
$x=3-h$
$\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0}[[3-h]]-[2(3-h)-1]=3-5=-2$
$\lim _{x \rightarrow 3} f(x)$ does exist.
Option $B : f(x)=[x]-x$ at $c=1$
Let's check right hand limit at $c=1$
$x=1+h$
$\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}[1+h]-(1+h)=1-1=0$
Let's check Left hand limit at $c=1$
$x=1-h$
$\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}[1-h]-(1-h)=0-1=-1$
$\lim _{x \rightarrow 1} f(x)$ does not exist.
Option $C : f( x )=\left( x ^{2}\right)-(- x )^{2},$ at $c =0$
Let's check right hand limit at $c=0$ $x=0+h$
$\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0}(h)^{2}--h^{2}=0-1=-1$
Let's check Left hand limit at $c=0$
$x=0-h$
$\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0}(-h)^{2}-h^{2}=0-0=0$
$\lim _{x \rightarrow 0} f(x)$ does exist.
Hence, Option $B$ and $C$.
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