Solution:

The provided function is

$f(x)= \begin{cases}\sin x-\cos x, & \text { if } \quad x \neq 0 \\ -1, & \text { if } \quad x=0\end{cases}$

Let us now find the L.H.L and R.H.L. at $x=0$.

At $x=0$,
Left Hand Limit
$=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} f(-h)$

$\Rightarrow \lim _{h \rightarrow 0} \sin (-h)-\cos (-h)=\lim _{h \rightarrow 0}(-\sin h-\cos h)=-0-1=-1$

Right Hand Limit $=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} f(h)$

$\Rightarrow \lim _{h \rightarrow 0} \sin (h)-\cos (h)=\lim _{h \rightarrow 0}(\sin h-\cos h)=0-1=-1$

And, here it is given $f(0)=-1$

Therefore, $\lim _{\lambda \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow D^{-}} f(x)=f(0)$

As a result, $f(x)$ is continuous at $x=0$