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

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.

$x^2$ 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:

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

• I've edited your answer to make use of $\LaTeX$. Please make sure it still represents your original intent. – apnorton May 16 '13 at 15:41

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 $f'(0)=0$.

$$\lim (f(x) g(x)) =\lim (f(x))\lim (g(x)),$$ 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.

• your logic is flawed. the equation you have is a means of finding limits, not a definition for when a limit exists. The use of the squeeze theorem is perfectly fine. If you haven't seen it before, you now have something new to read and learn about. – AsheKetchum Oct 12 '17 at 15:33