limit as norm goes to infinity

Question

Find the limit $\lim_{||(x,y)||\to \infty} \frac{x+y}{x^2-xy+y^2}$. If we go along a path y=0 the function goes to 0, so if the limit exists it must be 0.

That is we need to find a K so that for each $\epsilon$ if ||(x,y)||>K than $|\frac{x+y}{x^2-xy+y^2}|<\epsilon$.
$\frac{x+y}{x^2-xy+y^2}=\frac{(x+y)^2}{x^3+y^3}$

Answer 1

Let $(x,y) = (r\cos\theta, r\sin \theta)$, then \begin{align} \frac{x+y}{(x^2+y^2)-xy} = \frac{r\cos \theta + r \sin\theta}{r^2 - r^2\sin\theta\cos\theta} = \frac{1}{r}\frac{\cos\theta + \sin \theta}{1 - \tfrac12\sin(2\theta)} \end{align}

Now using basic inequality $-1\leq \cos\theta,\sin\theta \leq 1$, we obtain $$ -\frac{4}{3r} =\frac1r \frac{(-1)+(-1)}{1-\tfrac12(-1)} \leq \frac1r \frac{\cos\theta + \sin\theta}{1-\tfrac12\sin(2\theta)} \leq \frac1r \frac{(1)+(1)}{1-\tfrac12(1)} = \frac{4}{r}$$ The bounds are independent of $\theta$, so $$\lim_{r\to \infty}-\frac{4}{3r} \leq \lim_{r\to\infty}\frac1r \frac{\cos\theta + \sin\theta}{1-\tfrac12\sin(2\theta)} = \lim_{\|(x,y)\|\to\infty} \frac{x+y}{x^2+ - xy +y^2} \leq \lim_{r\to \infty} \frac{4}{r}$$

By the squeeze theorem:

$$ \lim_{\|(x,y)\|_2\to\infty} \frac{x+y}{(x^2+y^2)-xy} =\lim_{r\to\infty} \frac{1}{r}\frac{\cos\theta + \sin \theta}{1 - \tfrac12\sin(2\theta)} = 0$$

Answer 2

Hints: Show the inequality $t^2-t+1\geq (1+t^2)/2$ (for all $t\in \mathbb{R}$) and use it to show that for all $x,y$ we have $x^2-xy+y^2\geq \frac{x^2+y^2}{2}$, and use to finish that $|x|\leq \sqrt{x^2+y^2}$ and $|y|\leq \sqrt{x^2+y^2}$