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 $$\lim_{n\to\infty}\frac{x_n^2y_n}{3x_n^2-2x_ny_n+y_n^2}$$ 

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.
