Question

Check whether or not the function f, defined on R by

f(x) = { 3x^2sin(1/2x), when x≠0

{ 0 ,when x=0

is derivable on R. If it is, is f' continuous at x=0? If f is not derivable , then define a derivable function on R

Answer 1

Apply the Squeeze Theorem:

$$-1\le\sin(1/2x)\le1, x\in \R$$

Then

$$-3x^2\le3x^2\sin(1/2x)\le3x^2, x\in \R$$

We see that

$$\lim\limits_{x\to0}(-3x^2)=0=\lim\limits_{x\to0}(3x^2)$$

Then by the Squeeze Theorem

$$\lim\limits_{x\to0}(3x^2\sin(1/2x))=0$$

$$\lim\limits_{x\to0}(3x^2\sin(1/2x))=0=f(0)$$

The function $f(x)$ is continuous on $\R.$

Apply the Squeeze Theorem:

$$-1\le\sin(1/2h)\le1, h\in \R$$

Then

$$-3h\le3h\sin(1/2h)\le3h, h\in \R$$

We see that

$$\lim\limits_{h\to0}(-3h)=0=\lim\limits_{h\to0}(3h)$$

Then by the Squeeze Theorem

$$\lim\limits_{h\to0}(\dfrac{f(0+h)-f(0)}{h})=\lim\limits_{h\to0}(\dfrac{3h^2\sin(1/2h)}{h})$$

$$=\lim\limits_{h\to0}(3h\sin(1/2h))=0=f'(0)$$

The function $f(x)$ is derivable on $\R.$

$$f'(x)=6x\sin(1/2x)+3x^2\cos(1/2x)(-\dfrac{1}{2x^2})$$

$$=6x\sin(1/2x)-\dfrac{3}{2}\cos(1/2x)$$

$$\lim\limits_{x\to0}(\cos(1/2x))=\text{does not exist}$$

If $x_n=\dfrac{1}{4\pi n}$ then $\cos(1/2x_n)=1, n\to \infin$

If $x_n=\dfrac{1}{\pi+2\pi n}$ then $\cos(1/2x_n)=0, n\to \infin$

Therefore $\lim\limits_{x\to0}(f'(x))=\text{does not exist}.$

The function $f'(x)$ is not continuous on $\R.$