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

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 ?

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

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.