x sin(1/x) is continuous at 0

Builds on Squeeze theorem Continuity at a point

Example

Let $f(x) = x \sin(1/x)$ for $x \ne 0$ , and let $f(0) = 0$ . Then $f$ is continuous at $0$ , and so $f$ is continuous on $\mathbb{R}$ . Indeed, $|f(x)| \le |x|$ for all $x$ , so $f(x) \to 0$ as $x \to 0$ by the squeeze theorem.

Remarks

The factor $x$ damps the oscillation of $\sin(1/x)$ . The function is not differentiable at $0$ . The difference quotient $(f(x) - f(0))/x = \sin(1/x)$ has no limit. The function $x^2 \sin(1/x)$ is differentiable with a discontinuous derivative.

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Limit point of a set definition
Sequence definition
Limit of a function definition
Real sequence definition
Continuity at a point definition
Limit of a sequence definition
Squeeze theorem statement
x sin(1/x) is continuous at 0 example

Tag	`0EZMAY1FQD`
URI	`m2://0EZMAY1FQD`
Kind	example
Writing	W0 stub
MSC	26A15
Instance of	Continuous function
Uses	Squeeze theorem, Continuity at a point

x sin(1/x) is continuous at 0

Example

Remarks

Reading path

Further reading

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