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?

  • 1
    $\begingroup$ For $h$ sufficiently close to 0, $\lfloor\sin(\log(h+2))\rfloor=0$, so the derivative exists and equals 0. $\endgroup$ – David M. Jun 15 '18 at 14:55

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.


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