Sin is caught in the box

Let's consider the 2 limits:

$R = \displaystyle \lim_{x\rightarrow0^{+}} \frac{sinx}{x}$

$L = \displaystyle \lim_{x\rightarrow0^{-}} \frac{sinx}{x}$

In either case, $\frac{sinx}{x} < 1$ for $x= 0 \pm \delta$ . The ratio approaches the limiting value 1 as $x$ approaches 0. However, the ratio is never equal to 1.

$\displaystyle \Rightarrow \left\lfloor \displaystyle \lim_{x \rightarrow 0} \frac{sinkx}{x} \right\rfloor = \left\lfloor \displaystyle \lim_{x \rightarrow 0} k\times \frac{sinkx}{kx} \right\rfloor$

No doubt, the limiting value is $k$ but as such it never becomes equal to $k$ and stays smaller than $k$ .
Hence,

$\left\lfloor \displaystyle \lim_{x \rightarrow 0} k \times \frac{sinkx}{kx} \right\rfloor = k-1$

$\displaystyle \Rightarrow \displaystyle \sum_{k=1}^{10} \left\lfloor \displaystyle \lim_{x \rightarrow 0} \frac{sinkx}{x} \right\rfloor = \displaystyle \sum_{k=1}^{10} k-1 = 0 + 1 + \ldots + 9 = \boxed{45}$