$g(x)=\left\{\begin{array}{cl}\frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\1, & x=-1\end{array} \text { and } h(x)=2[x]-f(x),\right.$
where $[x]$ is the greatest integer $\leq x$. Then the value of $\lim _{x \rightarrow 1} g(h(x-1))$ is
$=\lim _{ k \rightarrow 0} g (-2+1) \because f ( x )=-1 \forall x < 0$
$= g (-1)=1$
$RHL =\lim _{ k \rightarrow 0} g ( h ( k )) , k > 0$
$=\lim _{ k \rightarrow 0} g (-1) , \because f ( x )=1, \forall x > 0$
$=1$
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$PROPERTY 1$ if $\lim _{ h \rightarrow 0} \frac{ f ( h )- f (0)}{\sqrt{| h |}}$ exists and is finite, and $PROPERTY 2$ if $\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h^2}$ exists and is finite.
Then which of the following options is/are correct ?
$(1)$ $f(x)=x|x|$ has $PROPERTY$ $2$ $(2)$ $f(x)=x^{2 / 3}$ has $PROPERTY$ $1$ $(3)$ $f(x)=\sin x$ has $PROPERTY$ $2$ $(4)$ $f(x)=|x|$ has $PROPERTY$ $1$