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?
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
$$g(x) = \left\{\begin{array}{l l} \sin(1/x) &amp; \qquad \text{if $x \neq 0$}\\
c &amp; \qquad \text{if $x=0$}       
\end{array}\right.$$
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 $\delta &gt; 0$ there exists $x_1,x_2 \in (-\delta,\delta) \setminus \{0\}$ such that $|f(x_1)-f(x_2)| &gt; 2\epsilon$. Let $N=\lceil \frac{1}{2\pi \delta} \rceil + 1$. It follows that $2\pi N + \frac{\pi}{2} &gt; \frac{1}{\delta}$ and $2\pi N +  \frac{3 \pi}{2} &gt; \frac{1}{\delta}$. 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 $|f(x_1)-f(x_2)|=2&gt;1=2\epsilon$ as desired. Now let $c$ be an element of $\mathbb{R}$ we have (by the triangle inequality) that $2\epsilon &lt;  |f(x_1) - f(x_2)| \leq |f(x_1) -c| + |f(x_2) - c|$ 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 $\epsilon &gt;0$ such that for all $\delta &gt;0$ there exists $x$ in $(-\delta,\delta)\setminus \{0\}$ such that $|f(x)-c|&gt;\epsilon$, 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 $a$.
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.