I was doing exercises on Howard Anton Calculus and I came across a problem which asks to find:

$$ \lim_{(x,y) \rightarrow (0,0)} (x^2+y^2)\sin\left( \frac{1}{x^2+y^2} \right).$$.

Intuitively we know the answer: $0$, but is there a step by step procedure that can be proposed as an argument ?

  • $\begingroup$ Of course: Bounded function (in some neighborhood of $\;(0,0)\;$ , in this case) times a function that converges to zero converges to zero. $\endgroup$
    – DonAntonio
    Jul 28, 2019 at 10:43

2 Answers 2


Hint. Note that the sine values stays in the bounded set $[-1,1]$, and therefore for $(x,y)\not=(0,0)$, $$0\leq \left|(x^2+y^2) \sin\left( \frac{1}{x^2+y^2} \right)\right|\leq x^2+y^2.$$


Another possible approach is to notice that the function $f(x,y)=x^2+y^2$ tends to $0$ as $(x,y)$ goes to $(0,0)$. Then, you can reduce yourself to the limit $$\lim_{z \to 0} z \sin(\frac{1}{z})$$ with the change of variable $z=f(x,y)$, and we know that the above limit is $0$. The nice thing is that this approach works any time you have a "disguised" known limit of one variable.


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