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I am trying to prove the following limits using the delta-epsilon method. Can you help me out?

1.$$ \lim_{(x,y)\to(2,3)}(3x^2y^2 + 4xy-12) = 120$$ 2.$$ \lim_{(x,y)\to(0,0)}\frac{5x^2y}{x^2+y^2} = 0$$

How do I work this out with the upper and lower boundaries?

guest315
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    For the first one, I would appeal to "product and sum of continuous functions is continuous". Else you may repeat the proof of that claim carefully with your particular functions. For the second one, I would use that $x^2$ is less than or equal to $(x^2+y^2)$ – Pedro Oct 11 '13 at 00:49
  • Choose (delta) = (epsilon/5) for the second problem. – Pratik Singhal Dec 01 '13 at 06:47

3 Answers3

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Hint:

(2)

$$ \Big|{5x^2y}\Big| \leq 5|x^2||y|\leq 5 (x^2+y^2)\sqrt{x^2+y^2}. $$

Note:

$$ |x|\leq \sqrt{x^2+y^2},\quad |y|\leq \sqrt{x^2+y^2} $$

Mhenni Benghorbal
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I always prefer to let my variables go to $0$, so, in $\lim_{(x,y)\to(2,3)}(3x^2y^2 + 4xy-12)$, let $x = 2+u$ and $y = 3+v$.

This expression becomes

$\begin{align} 3x^2y^2 + 4xy-12 &=3(2+u)^2(3+v)^2 + 4(2+u)(3+v)-12\\ &=3(4+4u+u^2)(9+6v+v^2) + 4(6+3u+2v+uv)-12\\ &= 3\cdot 4\cdot 9+12 + \text{ terms in }u\text{ and }v\\ &= 108+12 + \text{ terms in }u\text{ and }v\\ &= 120 + \text{ terms in }u\text{ and }v\\ \end{align} $

so the limit as $u, v \to 0$ is $120$.

Note that you do not need to know the exact ecpressions with $u$ and $v$ since they all go to $0$ with $u$ and $v$.

marty cohen
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    How does this use a delta-epsilon proof? – guest315 Oct 11 '13 at 01:36
  • A person, can do this without taking even this much pain , just plug in the numbers 2 and 3 for x and y. We can do this as the first function asked is continuous at the given point. – Pratik Singhal Dec 01 '13 at 06:37
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My hint:

$$L=\lim_{(x,\: y)\to (0,\: 0)}\frac{5x^2y}{x^2+y^2}=\lim_{(x,\: y)\to (0,\: 0)}\frac{xy}{x^2+y^2}.5x$$

Because: $$\left|\frac{xy}{x^2+y^2}\right|\leq \frac{1}{2}\to L=0$$

Iloveyou
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