Show that f(x) = $\sin(\frac1x)$ is not differentiable at $x=0$

Question

I've been looking forever and have yet to find any examples of someone actually working out the limit of this problem:

$$\lim_{x\to0} \sin(\frac1x)$$

I'm stuck at the beginning:

$$\lim_{h\to0} \frac{(\sin(\frac1{x+h})-\sin(\frac1x))}h$$

I assume I need to evaluate from the left and the right, but I'm not sure how to work it out past this point either way.

Answer 1

Easier: isn't continuous at zero because $\lim_{x\to 0}\sin(1/x)$ does not exists. Why? Take sequences $x_n\to 0$ and $y_n\to 0$ with $\sin(1/x_n) = 0$ and $\sin(1/y_n) = 1$.

Answer 2

Choose $x_n=\frac 1{2\pi n+\pi/2}$ and $x_n=\frac 1{2\pi n}$. both limits converge to 0. But pluging them into your function yields different limits. Hence, function is not continous at $x=0$, but this is necessary for differentiability.

Answer 3

First, you ususally define the function $f(x) = \begin{cases} \sin\frac{1}{x} & x \ne 0\\0&x=0\end{cases}$

Making f(x) defined at 0 makes the problem a little bit more difficult.

Suppose $f(x)$ is differentiable at 0.

Then this limit exists:

$\lim_\limits{h\to0} \frac{\sin\frac1{h}}h = L$

from the definition of limit:

$\forall \epsilon>0, \exists \delta>0$ such that $|h|<\delta \implies |\frac{(\sin(\frac{1}{h})}h - L| < \epsilon$

There exists an $|h|<\delta$ such that $\sin(\frac1{h}) = 1$, and an $|h|<\delta$ such that $\sin(\frac1{h})= -1$

$\forall \delta>0, \exists h$ such that $|h|<\delta$ and $|\frac{\sin(\frac1{h})}h - L| = \frac{1}{h}$

let $\epsilon$ < $\frac{1}{\delta}$

The limit does not exist.

Answer 4

A function can only be differentiable at a point $x$ if $x$ is in the domain of the function. $x=0$ is not in the domain of $\sin(1/x)$.

Answer 5

The function, as given, is not differentiable at $0$ because it is not even defined there. Thus, we generally assign $f(0) = 0$.

Note that it suffices to show $f$ is not continuous at $0$. Firstly, note that if $f$ is continuous at some point $c$ in its domain, there exists a $\delta$ such that for all $x,y \in (c-\delta, c+\delta) \setminus \{c\}$, $|f(x) - f(y)|<2\epsilon$. However, taking $\epsilon<1$ this fails. In every neighbourhood of $0$, $f(x) = \sin \frac 1x$ takes on values $-1$ and $1$.