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Show that $$ \lim\limits_{(x,y)\to(0,0)}\dfrac{x^2y^2}{x^2+y^2}=0$$ My try: We know that, $$ x^2\leq x^2+y^2 \implies x^2y^2\leq (x^2+y^2)y^2 \implies x^2y^2\leq (x^2+y^2)^2$$ Then, $$\dfrac{x^2y^2}{x^2+y^2}\leq x^2+y^2 $$ So we chose $\delta=\sqrt{\epsilon}$

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4 Answers 4

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Or alternatively, by AM-GM we get $${x^2+y^2}\geq 2|xy|$$ so $$\frac{x^2y^2}{x^2+y^2}\le \frac{x^2y^2}{2|xy|}=\frac{1}{2}|xy|$$ and this tends to zero if $x,y$ tend to zero.

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  • $\begingroup$ I am aware of this answer, because i just saw it in another question about the same limits. But i am only looking for answers that use the epsilon-delta definition. $\endgroup$
    – Varazda
    Commented Aug 16, 2019 at 18:21
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    $\begingroup$ @hamzaboulahia You can bound the RHS by $(x^2 + y^2)/4$ (see the first inequality), hence you can take $\delta = 2\sqrt{\epsilon}$. More directly, you can get the same bound by squaring both sides of $x^2 + y^2 \geq 2|xy|$ and dividing by $x^2 + y^2$. $\endgroup$
    – user169852
    Commented Aug 16, 2019 at 18:27
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In two variables the epsilon-delta definition for $\lim_{\substack{x\to a\\ y\to b}}f(x,y)=L$ means that for every $\epsilon >0$ there exists a $\delta>0$ such that $\big|f(x,y)-L\big|<\epsilon$ whenever $0<\sqrt{(x-a)^2+(y-b)^2}<\delta$.

In your case, you want to show that $\big|f(x,y)-0\big|<\epsilon$ whenever $0<\sqrt{x^2+y^2}<\delta$. You did this by showing that

\begin{align}x^2y^2\leq (x^2+y^2)^2\implies\bigg|\frac{x^2y^2}{x^2+y^2}-0\bigg|\le\bigg|\frac{(x^2+y^2)(x^2+y^2)}{x^2+y^2}\bigg|=\bigg|x^2+y^2\bigg|=x^2+y^2\end{align}

so that you could choose $\delta=\sqrt{\epsilon}$ and then get the required form of $\big|f(x,y)-0\big|<\epsilon$.

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HINT \begin{align*} 0\leq x^{2} \leq x^{2} + y^{2} \Longleftrightarrow 0\leq \frac{x^{2}}{x^{2}+y^{2}} \leq 1 \Longleftrightarrow 0\leq \frac{x^{2}y^{2}}{x^{2}+y^{2}}\leq y^{2} \end{align*}

Then apply the squeeze theorem.

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Tips

$\lim\limits_{\left(x,y\right)\rightarrow\left(0,0\right)} \dfrac{x^2 y^2}{x^2+y^2} = \lim\limits_{\left(x,y\right)\rightarrow\left(0,0\right)} \dfrac{1}{\frac{1}{x^2}+\frac{1}{y^2}}$

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