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Prove that: For all $\epsilon>0$ exist $\delta>0$ which depends on $\epsilon$, such that:

$$\left| {\frac{{2{x^2}y - x{z^2}}}{{yz - {z^2}}}}-0 \right|<\epsilon$$ ever that $$0 < \sqrt {{x^2} + {y^2} + {z^2}} < \delta $$

I find it very difficult to find $\delta$ in terms of $\epsilon$. Any suggestions to prove this?

$$\mathop {\lim }\limits_{(x,y,z) \to (0,0,0)} \left( {\frac{{2{x^2}y - x{z^2}}}{{yz - {z^2}}}} \right)=0$$

thanks.

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What is $\delta$ and $\epsilon$? And where is your actual question: In the title or in the body? –  draks ... Jul 24 '12 at 9:39
2  
Very hard to prove something that is false. So it is not surprising that you cannot find appropriate $\delta$: there isn't one. –  André Nicolas Jul 24 '12 at 9:43
    
your questions do not match . Please check once . –  Theorem Jul 24 '12 at 9:52
    
already clarified the question –  mathsalomon Jul 24 '12 at 9:53
    
Clarify the title! Is there a 2 missing? –  draks ... Jul 24 '12 at 11:21

1 Answer 1

up vote 2 down vote accepted

Hint: Consider sequences $$ \begin{align} (x_n,y_n,z_n)&=(n^{-1/2},2n^{-1},n^{-1})\\ (x_n,y_n,z_n)&=(0,2n^{-1},n^{-1}) \end{align} $$ then you get $$ \begin{align} \lim\limits_{n\to\infty}\frac{2x_n^2 y_n-x_nz_n^2}{y_n z_n-z_n^2}&=\lim\limits_{n\to\infty}(4-n^{-1/2})=4\\ \lim\limits_{n\to\infty}\frac{2x_n^2 y_n-x_nz_n^2}{y_n z_n-z_n^2}&=\lim\limits_{n\to\infty}0=0 \end{align} $$ Thus we conclude that the limit $$ \lim\limits_{(x,y,z)\to (0,0,0)}\frac{2x^2 y-x z^2}{y z-z^2} $$ doesn't exist.

P.S. I used approach from this answer.

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Norbert its generally easy to show that the limit doesn't exist compared to it exists. Is there a way to show that limit exists for multivariable case ? –  Theorem Jul 24 '12 at 9:43
    
In general it is hard to show that limit does exist. But in most cases you had to make some tricky estimations and use standard inequalities –  userNaN Jul 24 '12 at 10:01
    
@mathsalomon You are welcome! –  userNaN Jul 24 '12 at 10:02
1  
@Theorem: In 2D, we can take $r_\alpha(t)=(t,\alpha t)$ and put it into your function to find path wise limit of $f$ when $t$ tends to $0$. After simplifying the original function, if the limit of last expression approaches to zero then probably your original function has limit $0$ at $(0,0)$. Now, we have to use $\epsilon, \delta$ to prove your limit. –  Babak S. Jul 24 '12 at 10:22
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Ok I'll write in more detail –  userNaN Jul 24 '12 at 10:50

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