Proving a limit using epsilon/delta

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SUMMARY

The limit of the function \(\frac{xyz}{x^2+y^2+z^2}\) as \((x,y,z)\) approaches \((0,0,0)\) is proven to be 0 using the epsilon-delta method. The key relationship established is \(|\frac{xyz}{x^2+y^2+z^2}| \leq |\frac{x}{2}|\), which leads to the conclusion that \(|\frac{x}{2}| < \frac{\delta}{2} = \epsilon\). The inequality \(yz \leq \frac{1}{2}(y^2 + z^2)\) is crucial for simplifying the expression and establishing the limit. This method effectively demonstrates the limit's behavior in three-dimensional space.

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Homework Statement


Prove that \mathop {\lim }\limits_{(x,y,z) \to (0,0,0) } \frac{xyz}{{x^2+y^2+z^2 }} = 0
using the epsilon-delta method

The Attempt at a Solution


0&lt;|(x,y,z)-(0,0,0)|=\sqrt{x^2+y^2+z^2}&lt;\delta
Now I have to rewrite:
0&lt;\left|\frac{xyz}{{x^2+y^2+z^2 }}-0\right|&lt;\epsilon
So that I find a relationship between epsilon and delta.

This is where I get stuck... I can't figure out how to do that.

This is one of my attempts:
0&lt;\left|\frac{xyz}{{x^2+y^2+z^2 }}-0\right|\leq \left|\frac{xyz}{{x^2}}\right|=\left|\frac{yz}{{x}}\right|

Any help is very much appreciated!
 
Last edited:
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Hint: yz<=1/2*(y^2+z^2)
 
Then it is easy:

0&lt;\left|\frac{xyz}{x^2+y^2+z^2}-0\right|\leq\left|\frac{xyz}{y^2+z^2}\right|\leq\left|\frac{xyz}{2yz}\right|=\left|\frac{x}{2}\right|&lt;\frac{\delta}{2}=\epsilon

How did you come up with: yz<=1/2*(y^2+z^2) ?
(Why is it true at all?)
 

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