# Function continuity and differentiability

I am given:

$$f(x) = \begin{cases} x^{4/3}\sin(1/x) & \text{if x\neq0} \\ 0 & \text{if x=0} \\ \end{cases}$$

I need to show that

1. $f$ is cts on $\mathbb{R}$
2. $f$ is differentiable on $\mathbb{R}$
3. $f'(x)$ is unbounded on any open interval containing 0.

• Continuity and differentiability on $\mathbb{R}^*$ are obvious. For continuity at $0$, show $\lim_0 f=0$, and for differentiability at $0$, show $\lim_0 f(x)/x=0$. Each time using $|\sin |\leq 1$. For 3, try to plug $x_n=1/n\pi$ in the derivative. – Julien Apr 22 '13 at 3:16
• hmmm, i'm sorry, could you elaborate on that? – Peej Gerard Apr 22 '13 at 3:18
• For 1, can you see that $f$ is continuous on $\mathbb{R}^*$ and that continuity at $0$ amounts to $\lim_0f=0$? If so, what would you do to prove the latter? Hint: squeeze. – Julien Apr 22 '13 at 3:20
• I see what you mean for continuity, but how can I be sure that the fcn is differentiable everywhere (including 0)? – Peej Gerard Apr 22 '13 at 3:29
• It is clearly differentiable outside of $0$, right? Then at $0$, apply the definition and compute $\lim_0(f(x)-f(0)/(x-0)$. That is $\lim_0 f(x)/x$. Show that this is $0$, which by def means that $f$ is diff at $0$ with $f'(0)=0$. – Julien Apr 22 '13 at 3:30

$f$ is clearly continuous for $x\neq 0$ because $x^{4/3}\sin(1/x)$ is continuous. Since $$-x^{4/3}\leq x^{4/3}\sin(1/x)\leq x^{4/3}$$ by the squeeze thorem $$\lim_{x\rightarrow 0}x^{4/3}\sin(1/x) = 0$$ so $f$ is also continuous for $x=0$.
Similarly, $f$ is clearly differentiable at any $x\neq 0$, for $x=0$ use the definition of the derivative to get \begin{eqnarray*}f'(0)& =& \lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} \\&=& \lim_{h\rightarrow 0}\frac{f(h)}{h}\\ &=&\lim_{h\rightarrow 0}\frac{h^{4/3}\sin(1/h)}{h}\\ &=&\lim_{h\rightarrow 0} h^{1/3}\sin(1/h)\\ &=&0\end{eqnarray*}
so $f$ is differentiable at $x=0$ and we have $$f'(x) = \left\{\begin{array}{cc}4/3x^{1/3}\sin(1/x) + x^{4/3}\cos(1/x)(-\frac{1}{x^2}) & x\neq 0\\ 0 & x=0\end{array}\right.$$
Now notice that the term $|x^{4/3}\cos(1/x)(-\frac{1}{x^2})|$ grows unboundedly when $x$ goes to $0$. So the same is true for $f'(x)$.
The continuity and differentiability of $f$ on $\mathbb R-\{0\}$ follows immediately. Note, $$\lim_{x\to 0}f(x)=\lim_{x\to 0}\left(x^{4/3}\sin\frac{1}{x}\right)=0=f(0)$$ $$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\left(x^{1/3}\sin\frac{1}{x}\right)=0$$since $\lim_{x\to 0}x^{4/3}=\lim_{x\to 0}x^{1/3}=0$ and $\sin\frac{1}{x}$ is bounded in any deleted neighborhood of $0.$ Consequently $f$ is continuous and differentiable at $0.$