Let $f(x,y) = x^2 + xy + y$. Give an epsilon-delta proof that the $$\lim_{(x,y) \rightarrow (1,1)} f(x,y)=3.$$
I know that $|x-1| < \delta$ and that $|y-1| < \delta$. As well as $|xy|< (x^2 + y^2)/2$.
I start by trying to simplify $|x^2 + xy + y -3|$ in a way that makes $|y-1|$ and $|x-1|$ appear but I'm having a real difficulty.
Any help or hints with this solution would be greatly appreciated, it's been driving me nuts.
I sometimes find it's easier to shift the problem back to $(0, 0)$. If we let \begin{align*} u &= x - 1 \\ v &= y - 1 \end{align*} then substituting back into the function minus $3$ yields, $$(u + 1)^2 + (u + 1)(v + 1) + v + 1 - 3.$$ Expanding this expression yields, $$u^2 + uv + 3u + 2v.$$ We could try bounding $|u|$ and $|v|$ by $\delta$ from here, but we could also back-substitute $x$ and $y$: $$(x - 1)^2 + (x - 1)(y - 1) + 3(x - 1) + 2(y - 1).$$ Now you should be able to force each factor under $\varepsilon / 4$ by choosing an approrpiate $\delta$.