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Is there $\displaystyle \lim_{(x,y)\to (0,0)}\frac{x^2\sin^2(y)}{x^2+2y^2}=0$?

Idea : Let $\varepsilon> 0 $ be given. Let's do $\delta=\sqrt{\varepsilon}>0$. Then, if $0<||(x,y)||<\delta \ \Rightarrow \ 0<x^2+y^2< \varepsilon $, $$\begin{align} \Rightarrow & \quad x^2<x^2+y^2< \varepsilon \\ \Rightarrow & \quad |x^2\sin^2(y)| \leq x^2< \varepsilon \\ \Rightarrow & \quad |x^2\sin^2(y)| < \varepsilon \end{align}$$ But I do not know what else to do. This limit exists? Can you help me? Thank you.

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2 Answers 2

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Note that if $(x,y)\neq(0,0)$, then$$\left\lvert\frac{x^2\sin^2(y)}{x^2+2y^2}\right\rvert\leqslant\frac{x^2y^2}{x^2+y^2}$$and so it is enough to prove that$$\lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}=0.$$Take $\varepsilon>0$. Now, take $\delta=\sqrt\varepsilon$. If $\bigl\lVert(x,y)\bigr\rVert<\delta$, then $\bigl\lVert(x,y)\bigr\rVert=r$ for some $r<\delta$. Then $\lvert x\rvert\leqslant\sqrt{x^2+y^2}\leqslant r$ and $\lvert y\rvert\leqslant r$ too. So$$\frac{x^2y^2}{x^2+y^2}<\frac{r^4}{r^2}=r^2<\delta^2=\varepsilon.$$

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    $\begingroup$ It's clear that $x^2y^2<\delta ^4 $ but why you conclude that $\frac{x^2y^2}{x^2+y^2}<\frac{\delta ^4}{\delta ^2}$ if $x^2+y^2<\delta ^2$? $\endgroup$ Sep 11, 2019 at 22:48
  • $\begingroup$ @Yessit I've edited my answer. What do you think now? $\endgroup$ Sep 11, 2019 at 22:53
  • $\begingroup$ @JoséCarlosSantos Perfect. thank you $\endgroup$ Sep 11, 2019 at 23:01
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As José Carlos Santos pointed out, since $|\sin x| \leq |x|$,

$$\left\lvert\frac{x^2\sin^2(y)}{x^2+2y^2}\right\rvert\leqslant\frac{x^2y^2}{x^2+2y^2}$$

Now, $|x|=\sqrt{x^2}\leq\sqrt{x^2+2y^2}$, and similarly $|y|=\sqrt{y^2}\leq\sqrt{x^2+2y^2}$, thus

$$\frac{x^2y^2}{x^2+2y^2}\leq\frac{(x^2+2y^2)^2}{x^2+2y^2}=x^2+2y^2$$

and $\lim\limits_{(x,y)\rightarrow 0}x^2+2y^2 = 0$.

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