sin(1/x) has no limit at 0

Builds on Limit of a function Sequential characterization of limits

Counterexample

Let $f(x) = \sin(1/x)$ for $x \ne 0$ . Then $\lim_{x \to 0} f(x)$ does not exist, and neither one-sided limit exists. Let $x_n = 1/(2 \pi n + \pi/2)$ and $y_n = 1/(\pi n)$ . Both sequences tend to $0$ from the right, but $f(x_n) = 1$ and $f(y_n) = 0$ for all $n \ge 1$ .

Remarks

The argument uses the sequential characterization of limits. So no value of $f$ at $0$ makes $f$ continuous at $0$ . The point $0$ is an essential discontinuity. The graph of $f$ on $(0, 1]$ , with its closure, is the topologist's sine curve.

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Limit point of a set definition
Sequence definition
Limit of a function definition
Real sequence definition
Limit of a sequence definition
Sequential characterization of limits statement
sin(1/x) has no limit at 0 counterexample

Tag	`0AFWQZ1PKF`
URI	`m2://0AFWQZ1PKF`
Kind	counterexample
Writing	W0 stub
MSC	26A15
Uses	Limit of a function, Sequential characterization of limits

sin(1/x) has no limit at 0

Counterexample

Remarks

Reading path

Further reading

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