Why can't $\sin(1/x)$ be differentiated? I've read in textbooks that $\sin(\frac 1x)$ can't be differentiated because it oscillates too rapidly, but so what? Why does oscillation stop it from being differentiated?
 A: The function $f(x)=\sin(1/x)$ is trivially not differentiable at $0$ since it is not defined there. However we will prove that there is no possible value for $c$ so that
$$g(x) = \left\{\begin{array}{l l} \sin(1/x) & \qquad \text{if $x \neq 0$}\\
c & \qquad \text{if $x=0$}       
\end{array}\right.$$
is continuous at $0$ (and hence, see (A) below, can't be differentiable at $0$). To do so we will show that the $\lim_{x\to 0} f(x)$ does not exist. 
Proof: Let $\epsilon = \frac{1}{2}$. We will begin by showing that for each $\delta > 0$ there exists $x_1,x_2 \in (-\delta,\delta) \setminus \{0\}$ such that $|f(x_1)-f(x_2)| > 2\epsilon$. Let $N=\lceil \frac{1}{2\pi \delta} \rceil + 1$. It follows that $2\pi N + \frac{\pi}{2} > \frac{1}{\delta}$ and $2\pi N +  \frac{3 \pi}{2} > \frac{1}{\delta}$. This means that $x_1= \frac{1}{2\pi N + \frac{\pi}{2}}$, $x_2 = \frac{1}{2\pi N + \frac{3\pi}{2}}$ are in $(-\delta,\delta)\setminus \{0\}$. Since $f(x_1)=1$ and $f(x_2)=-1$ we see that $|f(x_1)-f(x_2)|=2>1=2\epsilon$ as desired. Now let $c$ be an element of $\mathbb{R}$ we have (by the triangle inequality) that $2\epsilon <  |f(x_1) - f(x_2)| \leq |f(x_1) -c| + |f(x_2) - c|$ so that at least one of $|f(x_1)-c|$ or $|f(x_2) -c|$ must be greater than $\epsilon$.
We have proved that for each $c$ in $\mathbb{R}$, there exists $\epsilon >0$ such that for all $\delta >0$ there exists $x$ in $(-\delta,\delta)\setminus \{0\}$ such that $|f(x)-c|>\epsilon$, and hence by definition that $\lim_{x\to 0} f(x)$ does not exist.
(A)
Let $f$ be a real valued function defined on an open interval containing $a$. If $f$ is differentiable at $a$, then $f$ is continuous at $a$.
Proof: Suppose $f$ is differentiable at $a$. Then by definition there exists $L$ such that $$\lim_{x\to a} \frac{f(x)-f(a)}{x-a}=L.$$
Using limit laws we have $$\lim_{x\to a} f(x) = \lim_{x\to a} f(x) -f(a) + f(a) = \lim_{x\to a} (f(x) -f(a)) + \lim_{x\to a} f(a) = \lim_{x\to a} \frac{f(x) -f(a)}{x-a}(x-a) + f(a)=\lim_{x\to a} \frac{f(x) -f(a)}{x-a}\lim_{x\to a}(x-a)+f(a)=L\times 0+f(a)=f(a)$$
and hence $f$ is continuous at $a$.
A: The function can be differentiated wherever it is defined: that is on the whole real line except $0$. At the point $0$, the derivative does not exist in the first place because the function is not defined there. If we extend the definition of the function by assigning a value to it at $0$, it still does not help in obtaining a derivative there, whatever value is chosen. This is essentially because of the oscillation that you mention, which grows ever wilder as $0$ is approached. From the definition of the derivative of a function $f$ at $0$, if it exists: $$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}.$$But the limit cannot exist in this case because, however $f(0)$ is chosen, the numerator will vary through a range, from $-1-f(0)$ to $1-f(0)$, and so does not tend to $0$, while the denominator does tend to zero. So the ratio oscillates over an ever wider range as $h$ tends to $0$. 
A: The function $f(x) = \sin(1/x)$ is differentiable on it's domain. The derivative is
$$
f'(x) = \cos(1/x)\frac{d}{dx}\frac{1}{x} = -\frac{1}{x^2}\cos\left(\frac{1}{x}\right).
$$
This derivative exists everywhere except at $x=0$. Note that the function $f(x)$ isn't defined at $0$. A function that is differentiable at a point is in particular defined at that point. So there is no problems with things oscillating to fast. 
A: Differentiation is rigourously defined by taking limits; because of the increasing rapidity of oscillation of $sin(1/x)$ as one approaches the origin, the limit that defines the derivative there doesn't converge. 
