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

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, \begin{align} \Rightarrow & \quad x^2 But I do not know what else to do. This limit exists? Can you help me? Thank you.

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$$\frac{x^2y^2}{x^2+y^2}<\frac{r^4}{r^2}=r^2<\delta^2=\varepsilon.$$

• 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 $x^2+y^2<\delta ^2$? Sep 11, 2019 at 22:48
• @Yessit I've edited my answer. What do you think now? Sep 11, 2019 at 22:53
• @JoséCarlosSantos Perfect. thank you Sep 11, 2019 at 23:01

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