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

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
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

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.
@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