Prove $\lim_{(x,y)\to (0,0)}\frac{xy\sin(y)}{x^2+y^2}$ using delta-epsilon argument I am trying to prove this limit using delta epsilon:
$$\lim_{(x,y)\to (0,0)}\frac{xy\sin(y)}{x^2+y^2}$$
I know how the individual components relate to delta etc but I can't put it together. Please help. (Alternatively, is it valid to use polar coordinates?)
 A: Hint. Since $\left|\frac{xy}{x^2+y^2}\right|\le1$, convince yourself that the limit in question is zero.
A: Given $\epsilon > 0$, take $\delta = \epsilon$. Then, if $|(x,y)| := \sqrt{x^{2}+y^{2}}\le \delta = \epsilon$, we have, obviously $|y|\le \sqrt{x^{2}+y^{2}} \le \epsilon$. Thus, using $\bigg{|}\frac{xy}{x^{2}+y^{2}}\bigg{|} \le 1$, we get:
$$\bigg{|}\frac{xy\sin(y)}{x^{2}+y^{2}}\bigg{|} \le |sin(y)| \le |y| \le \epsilon,$$
where I used $|\sin(x)| \le |x| $, as proved here.
A: we know that $\lim_{y \rightarrow 0} \sin y =0$ so there exists $\delta$ s.t $0<|y|<\delta \implies |\sin y| < \epsilon$
$$|\frac{\sin yxy}{x^2+y^2}|\leq |\sin y| < \epsilon$$
Note that $|y| \leq ||\textbf{x}|| < \delta $ hence above delta suffices
A: Note that we have $\left|\frac{xy}{x^2+y^2}\right|\le \frac12$ from the AM-GM inequality and that $|\sin(y)|\le |y|\le \sqrt{x^2+y^2}$.
Therefore, we have for any $\varepsilon>0$
$$\left|\frac{xy\sin(y)}{x^2+y^2}\right|\le \frac12 \sqrt{x^2+y^2}<\varepsilon$$
whenever $0<\sqrt{x^2+y^2}<\delta=2\varepsilon$.
