Hey I need to show that $\lim_{(x,y)\to(0,0)} {xy \over \sqrt{x^2 + y^2}} = 0.$ I'm not sure how to start manipulating this as I haven't gotten anything useful yet. Some help to get me going would be nice. Thanks
epsilon delta proof for 2 variable limit
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calculus
multivariable-calculus
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0I love when the question asked 5 years ago matches my homework exactly ... – 2018-09-05
2 Answers
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Notice that $|x| = \sqrt{x^2} \le \sqrt{x^2 + y^2}$ Using this fact, we have $\left\vert\frac{xy}{\sqrt{x^2 + y^2}}\right\vert = \frac{|x|}{\sqrt{x^2 + y^2}}|y|\le |y|$ which approaches $0$ as $(x,y)\rightarrow 0$.
It should not be too hard to convert this into a formal epsilon-delta proof.
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Let $f(x,y) = \dfrac{xy}{\sqrt{x^2 + y^2}}$ Set $x = r \cos(\theta)$, $y = r \sin(\theta)$, let $\sqrt{x^2 + y^2} = r$. $F(r,\theta) = f(r \cos(\theta), r \sin(\theta)) = \dfrac{r^2 \cos(\theta) \sin(\theta)}{r} = r \cos(\theta) \sin(\theta)$ $(x,y) \to (0,0) \implies r \to 0$ Now can you conclude what you want?
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0@William: But this is a very formal approach (+1). :) – 2012-09-26