Differentiability of $f(x) = x^2 \sin{\frac{1}{x}}$ and $f'$

Question

Let $f(x) = x^2 \sin{\frac{1}{x}}$ for $x\neq 0$ and $f(0) =0$.

(a) Use the basic properties of the derivative, and the Chain Rule to show that $f$ is differentiable at each $a\neq 0$ and calculate $f'(a)$.

You may use without proof that $\sin$ is differentiable and that $\sin' =\cos$.

Not even sure what this is asking.

(b) Show that $f$ is differentiable at $0$ and that $f'(0) =0$.

$\frac {f(x)-f(0)}{x-0} \to \lim_{x \to 0} x \sin(1/x)$.

$x \sin(1/x) \leq |x|$ and $\lim_{x \to 0} |x|=0$.

Thus $f(x)$ is differentiable at $0$; moreover $f^{'}(0)=0$.

(c) Show that $f'$ is not continuous at $0$.

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

In pieces: $\lim_{x \to 0} \cos (1/x)$.

$f^{'}(0-)$ nor $f{'}(0+)$ exists as $x \to 0$ $f^{'}(x)$ oscillates infinity between $-1$ and $1$ with ever increase frequency as $x \rightarrow 0$ for any $p>0$ $[-p,0]$, $[-p,p]$ or $[0,p]$ $f$ is not continuous.

Question: How to show more rigorously?

Answer 1

$x^2$ is continuous and differentiable over $\mathbb{R}$

$\sin(x)$ is continuous and differentiable over $\mathbb{R}$

$\frac 1 x$ is continuous and differentiable over all $\mathbb{R}$ except $0$. And is a function of $\mathbb{R \to R}$

$\displaystyle \sin\left(\frac 1 x\right)$ is therefore continuous and differentiable for all $\mathbb R$ except $0$ where it is undefined.

$\displaystyle x^2\sin\left(\frac 1 x\right)$ is therefore continuous and differentiable for all $\mathbb R$ except possibly $0$.

To compute $f'(a)$ use product rule followed by chain rule to find:

$$F'(a) = 2a\sin\left(\frac 1 a\right) - \cos\left(\frac 1 a\right)$$

Answer 2

Part (b). The function $f$ is differentiable at $0$ and has $f'(0)$ equal to the limit if the following limit exists: \begin{align} \lim_{x \to 0} \dfrac{f(x) - f(0)}{x-0} & = \lim_{x \to 0} \dfrac{f(x) - 0}{x} & \textrm{ as } f(0) = 0 \\ & = \lim_{x \to 0} \dfrac{x^2 \sin\left(\frac{1}{x}\right)}{x} & \\ & = \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) & \end{align}

Now we can use the Squeeze Theorem. As $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$, we have that $$0 = \lim_{x \to 0} x \cdot -1 \leq \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) \leq \lim_{x \to 0} x \cdot 1 = 0$$

Therefore, $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0$ and we have $f'(0)=0$.

Answer 3

$$\lim (f(x) g(x)) =\lim (f(x))\lim (g(x)),$$ provided that both limits exists.

In above case, lim $\sin (1/x)$ does not exist as $x$ tends to $0$. Therefore, the above method is flawed.

It is however true that $\lim x \sin(1/x) = 0$ as $x$ tends to $0$, but, the method of proving this result, as shown above, is not correct.

One need to use epsilon-delta method (involving rigorous maths) to prove this result.