I was trying to evaluate $$ \lim_{(x,y) \rightarrow (0,0) } \frac{x^2y^2}{x^2 + y^2} $$ but only partially managed to do so through implying $y=kx$ but didn't find any other appropriate functions for this.
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If you are familiar with the well-known inequality: $x^2+y^2\geq 2|xy|$, then you can use it: $0\leq \dfrac{x^2y^2}{x^2+y^2} = \left|\dfrac{xy}{2}\cdot \dfrac{2xy}{x^2+y^2}\right|=\dfrac{|xy|}{2}\cdot \dfrac{2|xy|}{x^2+y^2}\leq \dfrac{|xy|}{2}$, and since $|xy| \to 0$ when$x, y \to 0$, the squeeze theorem says the limit is $0$.
DeepSea
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schwartz cuachy? – Bak1139 Apr 04 '15 at 18:00
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Its am-gm ineq. – DeepSea Apr 04 '15 at 18:38
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See also here for the inequality used in this answer. – Martin Sleziak Sep 16 '15 at 13:35
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Another hint $$ \frac{x^2y^2}{x^2 + y^2} = \frac{1}{\displaystyle \frac{1}{x^2}+\frac{1}{y^2}} $$
GEdgar
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Hint. By the change of variables $$ x=r \cos \theta,\quad y=r \sin \theta, $$ you may observe that $$ \left|\frac{x^2y^2}{x^2 + y^2}\right|\leq r^2\cos^2 \theta\sin^2 \theta\leq r^2. $$
Olivier Oloa
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We have $$\left|\frac{x^2y^2}{x^2 + y^2}\right|\leq \left|\frac{x^2y^2}{y^2}\right|=x^2$$ Since $f(x,y)$ is between $0$ and $x^2\rightarrow 0$, the function is arbitrarily close to zero, when the distance between $(x,y)$ and $(0,0)$ is sufficiently small. It follows that the limit of $f$ is $0$, just to limit definition.
Mark
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A problem I see with this answer is that $\frac{x^2y^2}{y^2}$ is undefined for some values from the domain of the function $f(x,y)=\frac{x^2y^2}{x^2+y^2}$. – Martin Sleziak Sep 16 '15 at 13:36
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We have that
$$0\le\frac{x^2y^2}{x^2 + y^2}\le\frac{x^2y^2+y^4}{x^2 + y^2}=\frac{y^2(x^2+y^2)}{x^2 + y^2}=y^2\to 0$$
user
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