Limit of non-linear multi-variable function I'm trying to prove the limit of the following function is $0$:
$\lim_{(x,y) \to (1,-1)} {x^3} - {2xy^2} + 1$
I know that I'm trying to find a $\delta$ s.t $ 0 < \sqrt{(x - 1)^2 + (y + 1)^2} < \delta $ which implies $|{x^3} - {2xy^2} + 1| < \epsilon$ 
I tried factoring $x$ so that $|x({x^2} - {2y^2}) + 1|$ and then adding 1 and - 1 using the triangle inequality to try to simplify into the expression I wanted, but I don't seem to be getting anywhere. 
I think I should try to get $|\sqrt{y^2}|$ so I could say it's less than $|\sqrt{ (x - 1)^2 + (y + 1)^2}|$, and subsequently define what $\delta$ is, but I'm just having a hard time how I could factorize this with the $+1$ in the problem. Any suggestions?
 A: A polynomial $f(x)$ is a continuous function, and hence $\lim_{x\rightarrow a} f(x) = f(a)$ by the definition of continuity. Here $x,a\in\mathbb{R}^n$. If $a=(1,-1)$, then $f(a)$ in this case is just $0$.
If your job is to show that any polynomial is continuous, I suggest you show it for a constant function, and then for the function $f(x)=x$, and then use various results about sums and products of continuous functions being continuous.
Edit: I will add some advice if you want to show this limit directly. If you like, you can do a change of coordinates  where $z=x-1$ and $w=y+1$ so that your limit is to the origin. This might make things easier for you. We can notice that $z^2 \leq z^2+w^2$, and after taking square roots, we get $|z|<\delta$ if $\sqrt{z^2+w^2}<\delta$. Similarly, for $w$. Now that you have a bound on $z,w$ individually, you can bound any polynomial in $z$ and $w$ using the triangle inequality. For example, $|z^3+w^2+2w|\leq |z|^3+|w|^2+2|w| < \delta^3+\delta^2+2\delta$ which is less than $\epsilon$ for appropriately chosen $\delta$.
A: We have $$\begin{align}|x^3-2xy^2+1|&=|(x-1+1)^3-2(x-1+1)(y+1-1)^2+1|\\&=\left|(x-1)^3+3(x-1)^2+3(x-1)-2(x-1)\left[(y+1)^2+2(x+1)\right]-2\left[(y+1)^2-2(y+1)\right]\right|\end{align}$$

Idea: We have those $x-1$ and $y+1$ in the condition $\sqrt{(x-1)^2+(y+1)^2}<\delta$, we make them appear in the expression $x^3-2xy^2+1$ by writing $x$ as $x-1+1$ and $y$ as $y+1-1$.

If we choose $\delta=\epsilon/19$ we get 
$$|x-1|<\epsilon/19$$
and 
$$|y+1|<\epsilon/19$$
from where
$$\begin{align}&\left|(x-1)^3+3(x-1)^2+3(x-1)-2(x-1)\left[(y+1)^2+2(x+1)\right]-2\left[(y+1)^2-2(y+1)\right]\right|\\&\leq|x-1|^3+3|x-1|^2+3|x-1|+2|x-1|\left[|y+1|^2+2|x+1|\right]+2\left[|y+1|^2+2|y+1|\right]\\&\leq 19\epsilon/19=\epsilon\end{align}$$
