Exercise 864Calculus › Multivariable calculus › ContinuityL1Added 3 Nov 2026

A discontinuous function with bounded values

#discontinuity#bounded#counterexample

Statement

Show that the function
$f(x,y)=\sin\!\left(\frac{1}{x^2+y^2}\right) \, \text{for} \, (x,y)\neq(0,0), \, \, f(0,0)=0$
is not continuous at the origin.

Hint

Test the path $(x,0)$ and analyze the oscillation.

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Along $(x,0)$ . $f(x,0)=\sin(1/x^2)$ oscillates in $[-1,1]$ with no limit.

Conclusion. $f$ has no limit at $(0,0)$ ; not continuous there.