# 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.

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 $f'(0)=0$.

$x^2$ 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. May 16, 2013 at 15:41

$$\lim (f(x) g(x)) =\lim (f(x))\lim (g(x)),$$ 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. Oct 12, 2017 at 15:33