# For which values of $x_0\in\mathbb{R}$ is $f$ differentiable

(i) Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be given by \begin{equation} f(x)=\begin{cases} (x-1)^3\sin(\frac1{x-1}) & x\neq1 \\ 0 & x=1\end{cases} \end{equation}

Determine those $x_0\in\mathbb{R}$ for which $f$ is differentiable at $x_0$. Give the value of $f'(x_0)$ when it exists, and prove any assertions that you make. If $f'(x_0)$ does not exist, you must prove why.

(ii) For the function $f$ given in part (i), determine whether the statement $f'(1)=\lim_{x\rightarrow1}f'(x)$ is true or false. Justify your answer.

For i) I first set $x_0\in{\mathbb{R}-{0}}$ and then tried calculating the limit $\lim_{h\rightarrow0}\frac{f(x_0+h)-f(x_0)}h$ but it turns out to be very dirty and I can't really find what the limit is. It seems to be infinity. Also we cannot use the L'Hopital Rule.

For $x_0 \neq 1$ the problem is simple: at such points you can just use the usual rules of calculus to check that the derivative exists. For $x_0=1$ you need to study

$$\lim_{x \to 1} \frac{f(x) - f(1)}{x-1} = \lim_{x \to 1} \frac{(x-1)^3 \sin \left ( \frac{1}{x-1} \right ) - 0}{x-1} = \lim_{x \to 1} (x-1)^2 \sin \left ( \frac{1}{x-1} \right ).$$

Now how can you compute this last limit? If you don't see it, here's a hint: what are the behaviors of $(x-1)^2$ and $\sin \left ( \frac{1}{x-1} \right )$ separately as $x \to 1$?

From there you'll do something very similar to compute $\lim_{x \to 1} f'(x)$.

• what definition did you use for the first limit? Don't you use the definition of the derivative ? – snowman Nov 22 '14 at 17:44
• @snowman That is the definition of the derivative. You can write it as $\lim_{h \to 0} \frac{f(1+h)-f(1)}{h}$ if you want, but they're the same. – Ian Nov 22 '14 at 17:45
• but for when $x\neq1$, isn't x a variable? you made it 1... – snowman Nov 22 '14 at 17:48
• @snowman: since $(x-1)^2\to0$ and $\left|\sin\left(\frac1{x-1}\right)\right|\le1$, the limit of their product is $0$. If you are still not convinced, use $$-(x-1)^2\le(x-1)^2\sin\left(\frac1{x-1}\right)\le(x-1)^2$$ and the Squeeze Theorem. – robjohn Nov 22 '14 at 18:04
• You can, but since $\frac1{x-1}$ is differentiable for $x\ne1$ and $(x-1)^3$ and $\sin(x)$ are differentiable everywhere, you can use the chain and product rules. That is, unless your instructions were to use only the definition of the derivative. – robjohn Nov 22 '14 at 19:18