$\mathop {\lim }\limits_{x \to {a^ + }} {{\left\{ x \right\}\sin \left( {x - a} \right)} \over {{{\left( {x - a} \right)}^2}}}$ 

is equal to (where {.} denotes the fraction
part of x and $a \in N$

Solve :

The correct option is B 1

Put $\mathrm{x}=\mathrm{a}+\mathrm{h}$ and $\mathrm{h} \rightarrow 0$, then we get,

$\lim _{h \rightarrow 0} \frac{h \sin (a+h-a)}{(a+h-a)^{2}} \ldots(a+h=h$ because it is a fractional part function and $a$ is natural number)

$\begin{array}{l} =\lim _{h \rightarrow 0} \frac{\sin h}{h} \\ =1 \end{array}$

So, option B is correct.