1
$\begingroup$

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.

  • 0
    Clar$i$fy the title! Is there $a$ 2 missing?2012-07-24

1 Answers 1

2

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.

  • 0
    @norbert. Thanks2012-07-24