Is the function $f(x)=\lfloor\sin(\ln(x+2)\rfloor$ Differentiable at $x=0$ where $ \lfloor .\rfloor$ represents floor function.
I proved that it is continuous at $x=0$.
Since $$\lim_{x \to a} \lfloor(g(x)\rfloor=\lfloor\lim_{x \to a} g(x) \rfloor$$ when $\lim_{x \to a} g(x)$ is not an integer.
Now since $\sin(\ln 2)$ is not an integer we have
$$\lim_{x \to 0} \lfloor\sin(\ln(x+2)\rfloor=f(0)$$
Hence $f$ is continuous.
Now checking the right hand derivative we have
$$f'(0)=\lim_{h \to 0}\frac{\lfloor\sin(\ln(h+2)\rfloor}{h}$$
Now this limit is in indeterminate form. Can I know how to evaluate this limit?
Letting $g(x) =\sin(\ln(x+2))$. Because $0<\ln(2)<\pi/2$, we know $0<g(0)=\sin(\ln(2))<1$.
Furthermore, $g(x)$ is continuous for $x>-2$, in particular it's continuous at $x=0$.
Hence, there is some neighbourhood around $x=0$ where $0<g(x)<1$.
Hence, there is some neighbourhood around $x=0$ where $f(x)=\lfloor g(x) \rfloor =0$ (constant). Hence $f(x)$ is differentiable at $x=0$ - and the derivative is zero.