# Thread: Difficult epsilon proof question

1. ## Difficult epsilon proof question

Once again, the textbook from which this question comes from has no prior example which I could reference to even possibly answer this.

If $\displaystyle \epsilon>0$, show that $\displaystyle |2x^2-6xy+5y^2|<\epsilon$ when $\displaystyle (x^2+y^2)^{1/2}<(\epsilon/13)^{1/2}$.

My Prof. is a true sadist...

2. Through other online sources, I've come up with this so far. I suspect, however, that it has some pieces missing.

If we square $\displaystyle (x^2+y^2)^{1/2}<(\epsilon/13)^{1/2}$, we get $\displaystyle x^2+y^2<\epsilon/13$.
This says that $\displaystyle (x,y)$ lies inside a circle of radius $\displaystyle \sqrt{\epsilon/13}$ around the origin.
Using the triangle inequality, (which is $\displaystyle |a+b|\leq|a|+|b|$)
$\displaystyle |2x^2-6xy+5y^2|\leq |2x^2|+|6xy|+|5y^2|$
The condition $\displaystyle x^2+y^2<\epsilon/13$ means that separately, $\displaystyle x^2$, $\displaystyle y^2$ and $\displaystyle |xy|$ are all less than $\displaystyle \epsilon/13$.
Therefore, $\displaystyle |2x^2|+|6xy|+|5y^2|<(2+6+5)(\epsilon/13)$.

If there are any errors in this, I'd appreciate it if they were pointed out.

3. Seems correct to me.