# How to demonstrate the limit $\lim_{(x,y) \rightarrow (0,0)} \frac {x\sin(xy)}{x^2+y^2}$ with $\epsilon$ and $\delta$!

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

I check with different paths to know the limit as pero succesions in both cases the result was 0.

If I try the answer in Wolfram Alpha say the limit do not exist (all paths I try are equal to 0) but if I graph the function don't see any hole or discontinuity.

I use a delta, epsilon probe but I reach a part that not always the inequity is true.

Please help me solve it by epsilon, delta approach or probe the limit does not exist. Thanks.

Hint. First, note that $\,|\sin t|\le |t|$, for all $t$.

Hence $$\left|\frac{x\sin (xy)}{x^2+y^2}\right|\le \frac{x^2|y|}{x^2+y^2}\le |y|.$$

Using Yiorgos hint:

Let $\epsilon \gt 0$ be given. Choose $\delta = \epsilon:$

$(x^2+y^2)^{1/2} \lt \delta$

$\rightarrow$

$|\dfrac{x\sin(xy)}{x^2+y^2}| =$

$|\dfrac{x^2y}{x^2+y^2} \dfrac{\sin(xy)}{xy}| \le$

$|\dfrac{x^2y}{x^2+y^2}|×1 \le |y| \le$

$(x^2+y^2)^{1/2} \lt \delta = \epsilon$.

• Thanks!!! That's the way I probe it – Rodrigo Ramírez Sep 3 '17 at 19:46
• Welcome. Looks so simple,:) – Peter Szilas Sep 4 '17 at 22:48