1
$\begingroup$

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?

$\endgroup$
  • 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.