y = \sin\dfrac{1}{x}

Graph of the Function $y = \sin\dfrac{1}{x}$

$y = \sin\dfrac{1}{x}$ is the sine of a reciprocal. As $x$ approaches $0$ the quantity $\dfrac{1}{x}$ grows without bound, so infinitely many oscillations are packed into every neighborhood of the origin, however small. It is the standard textbook example of a limit that does not exist.

Domain and range

The domain is $x \neq 0$ and the range is the interval $[-1, 1]$ . Since $\dfrac{1}{x}$ takes every real value except $0$ , the sine attains all of its values.

Symmetry

From $f(-x) = \sin\left( -\dfrac{1}{x} \right) = -f(x)$ the function is odd, with rotational symmetry about the origin.

Zeros, crests and troughs

We have $\sin\dfrac{1}{x} = 0$ when $\dfrac{1}{x} = n\pi$ , that is at $x = \dfrac{1}{n\pi}$ .

$n$	Zero at $x = \dfrac{1}{n\pi}$
$1$	$\approx 0.3183$
$2$	$\approx 0.1592$
$3$	$\approx 0.1061$
$4$	$\approx 0.0796$

The zeros crowd endlessly toward the origin. The value $\pm 1$ is reached at $x = \dfrac{2}{(2n+1)\pi}$ , the rightmost crest being $\left( \dfrac{2}{\pi}, 1 \right) \approx (0.637, 1)$ .

Why the limit fails to exist

Consider the limit as $x \to 0$ . Two sequences tending to $0$ head for different destinations.

\begin{align*} x_n &= \frac{2}{(4n+1)\pi} \to 0, \quad \sin\frac{1}{x_n} = 1 \\ x_n' &= \frac{2}{(4n+3)\pi} \to 0, \quad \sin\frac{1}{x_n'} = -1 \end{align*}

The limit as $x \to 0$ therefore does not exist, and neither one-sided limit exists either.

What kind of discontinuity

The values stay between $-1$ and $1$ , so the $y$ -axis is not a vertical asymptote.

Function	Behavior as $x \to 0$	Kind of discontinuity
$\dfrac{1}{x}$	diverges	infinite
$\arctan\dfrac{1}{x}$	one-sided limits $\pm\dfrac{\pi}{2}$	jump
$\sin\dfrac{1}{x}$	no one-sided limits	oscillatory

Of the classification of discontinuities, this is the most intractable kind¹.

Behavior on the right

The derivative is as follows.

y' = -\frac{\cos\dfrac{1}{x}}{x^2}

For $x > \dfrac{2}{\pi}$ we have $\dfrac{1}{x} < \dfrac{\pi}{2}$ , so $\cos\dfrac{1}{x} > 0$ and $y' < 0$ : the function decreases steadily. As $x \to \infty$ we get $\dfrac{1}{x} \to 0$ and hence $y \to 0$ , making the $x$ -axis a horizontal asymptote. The function is placid far out and wilder the closer one comes to the origin.

What no graph can show

The number of oscillations in the interval $(0, a]$ is about $\dfrac{1}{2\pi a}$ , which grows without bound as $a$ shrinks. No amount of sampling can therefore render the neighborhood of the origin correctly. Every time the view is magnified new oscillations appear, and the scene never settles.

A link to topology

Adding the segment $-1 \leq y \leq 1$ of the $y$ -axis to the following set produces the topologist's sine curve².

\left\{\left(x, \sin\frac{1}{x}\right) : 0 < x \leq 1\right\}

It is the classic example of a set that is connected but not path-connected. It looks joined up, and yet no path along the curve ever reaches the origin.

Classification of discontinuities, Wikipedia
Topologist's sine curve, Wikipedia

Graph of the Function y=sin⁡1xy = \sin\dfrac{1}{x}y=sinx1​