Another limit from a math contest $\lim_{n\to\infty}\frac{x_n^2y_n}{3x_n^2-2x_ny_n+y_n^2}$ Let $(x_{n})_{n\ge1}$, $(y_{n})_{n\ge1}$ be real number sequences and both converge to $0$.
Evaluate $$\lim_{n\to\infty}\frac{x_n^2y_n}{3x_n^2-2x_ny_n+y_n^2}$$ 
 A: By the AM-GM, $|3t+1/t|\ge2\sqrt3$. Therefore, $|3t-2+1/t|\ge2\sqrt3-2$.
Thus, we have
$$
\begin{align}
\left|\frac{x_n^2y_n}{3x_n^2-2x_ny_n+y_n^2}\right|
&=\left|\frac{x_n}{3(x_n/y_n)-2+(y_n/x_n)}\right|\\
&\le\frac{|x_n|}{2\sqrt3-2}
\end{align}
$$
Therefore, as long as $\lim\limits_{n\to\infty}x_n=0$,
$$
\lim_{n\to\infty}\frac{x_n^2y_n}{3x_n^2-2x_ny_n+y_n^2}=0
$$
A: The proof suggested in the post is incomplete since the sequence $(z_n)_n$ could have many behaviours different of the two you cite. What is crucial here is that the denominator never goes close to zero. More quantitavely, note that, for every $(x_n,y_n)$,
$$
3x_n^2-2x_ny_n+y_n^2\geqslant|x_ny_n|.
$$
Hence, if $(x_n,y_n)=(0,0)$, the $n$th ratio $r_n$ is undefined. If $x_ny_n=0$ but $(x_n,y_n)\ne(0,0)$, then $r_n=0$. And if $x_ny_n\ne0$, then $|r_n|\leqslant|x_n|$. Thus, if $(x_n,y_n)\ne(0,0)$ for every $n$ and $x_n\to0$, then $r_n$ is well defined for every $n$ and $r_n\to0$. 
A: Your use of $\lim z_n$ is suspicious. Note that $z_n$ can be any sequence of positive numbers (just start with arbitrary $z_n$ and let $x_n=\frac1n, y_n=\frac{x_n}{z_n}$ if $z_n\ge1$ and $y_n=\frac1n, x_n=z_ny_n$ if $z_n<1$).
Note that $3x_n^2-2x_ny+y_n^2=2x_n^2+(x_n-y_n)^2\ge2x_n^2$, hence
$$ \left|\frac{x_n^2y_n}{x_n^2-2x_ny_n+y_n^2}\right|\le\left|\frac{x_n^2y_n}{2x_n^2}\right|= \frac{|y_n|}2,$$
which shows that $$ \lim_{n\to\infty}\frac{x_n^2y_n}{x_n^2-2x_ny_n+y_n^2}=0.$$
A: I don't know how many of your alters (see http://en.wikipedia.org/wiki/United_States_of_Tara ) know Lagrange multipliers, so I will give a bit extra. $3 x^2 - 2 x y + y^2$ is a positive definite quadratic form. Setting it equal to a positive constant gives and ellipse, a bounded figure. As a result, the function $x^2$ is bounded on such an ellipse, and Lagrange Multipliers tells us that $$ x^2 \leq \frac{1}{2} \left( 3 x^2 - 2 x y + y^2 \right)  $$ with the optima occurring where $x=y.$ As a result, except at the origin itself,
$$ \frac{x^2}{3 x^2 - 2 x y + y^2} \leq \frac{1}{2}, $$ and
$$ \left| \frac{x^2 y}{3 x^2 - 2 x y + y^2} \right| \leq  \left| \frac{y}{2} \right| . $$  
A little late now. I see that Andre had the better presentation, then deleted it in anguish over not having quite been first. So, as seen elsewhere,
$$  3 x^2 - 2 x y + y^2 = 2 x^2 + (x-y)^2 \geq 2 x^2. $$ Completing the square the other way gives
$$  3 x^2 - 2 x y + y^2 = \frac{1}{3} \left( 9 x^2 - 6 x y + 3 y^2 \right) =  \frac{1}{3} \left( (3x-y)^2 + 2 y^2 \right) \geq  \frac{1}{3} \cdot 2 y^2, $$ and
$$  3 x^2 - 2 x y + y^2 \geq \frac{2}{3}  y^2. $$
So, if either of $x,y$ is nonzero, the denominator is also nonzero and positive. That's called positive definite.
A: I hope I'm not duplicating, but this seems a bit different.  Rewrite the fraction as
$$\frac{y_n}{3 - 2 \frac{y_n}{x_n} + \left ( \frac{y_n}{x_n} \right )^2}$$
There are three cases:
1) $\lim_{n \rightarrow \infty} (y_n/x_n) = 0$.  Then the limit is $(1/3) \lim_{n \rightarrow \infty} y_n = 0$.
2)  $\lim_{n \rightarrow \infty} (y_n/x_n) = L \in (0,\infty)$.  Then the limit is 
$$\lim_{n \rightarrow \infty} \frac{y_n}{3 - 2 L + L^2} = 0$$
3) $\lim_{n \rightarrow \infty} (y_n/x_n) = \infty$.  Then the denominator of the fraction goes to zero, and therefore the limit is zero because the numerator also vanishes in the limit.
