The left-hand derivative of $f(x)=[x] \sin \pi x$ at $x=k$, $k$ being an integer, is

$(-1)^k(k-1) \pi$

We have,

$f(x)=[x] \sin \pi x$

∴ (LHD at x = k) = $\lim\limits_{x \rightarrow k^{-}} \frac{f(x)-f(k)}{x-k}$

⇒ (LHD at x = k) = $\lim\limits_{h \rightarrow 0} \frac{f(k-h)-f(k)}{-h}$

⇒ (LHD at x = k) = $\lim\limits_{h \rightarrow 0} \frac{[k-h] \sin (\pi k-\pi h)-[k] \sin \pi k}{-h}$

⇒ (LHD at x = k) = $\lim\limits_{h \rightarrow 0} \frac{(-1)^{k-1}[k-h] \sin \pi h-0}{-h}$

⇒ (LHD at x = k) = $\lim\limits_{h \rightarrow 0}(-1)^k[k-h] \frac{\sin \pi h}{h}$

⇒ (LHD at x = k) = $(-1)^k(k-1) \lim\limits_{h \rightarrow 0} \frac{\sin \pi h}{h}=(-1)^k(k-1) \pi$