How to prove the existence of the limit？

Question

Prove：$\quad\lim_{x\to0\\y\to0}\frac{x^2+y^2}{|x|+|y|}=0$

I don't know how to shrink $\lim_{x\to0\\y\to0}\frac{x^2+y^2}{|x|+|y|}$

Answer 1

$$\frac{x^2+y^2}{|x|+|y|} =\frac{x^2}{|x|+|y|} + \frac{y^2}{|x|+|y|} \le \frac{x^2}{|x|} + \frac{y^2}{|y|} = |x|+|y|$$

Answer 2

I'll forego the image: $$\lim_{x,y\to0} {x^2+y^2\over|x|+|y|}$$ We can write absolute values like so: $|x| = \sqrt{x^2}$. Secondly, because it is a two-dimensional limit, and we can easily work with the squares, I'll convert to polar coordinates. Recall that $x=r\cos(\theta)$ and $y=r\sin(\theta)$, and $r^2=x^2+y^2$. Finally, if both x and y are going to 0, that is, both are going toward the origin, then we can say r is going to 0.
Converted, we have $$\lim_{r\to 0} \frac{r^2}{\sqrt{r^2\cos^2(\theta)}+\sqrt{r^2\sin^2(\theta)}}$$. If we pull a common $\sqrt{r^2}$ out of the denominator, and convert the numerator to $\sqrt{r^4}$, we will have $$\lim_{r\to0} \sqrt{r^4\over r^2}\left({1 \over |\cos|+|\sin|}\right)$$
The radical will simplify to $\sqrt{r^2}$ = $|r|$. The limit of the product is equal to the product of the limits: $$\lim_{r\to0}|r|\cdot\lim_{r\to0}\Bbb{(\theta \,stuff)} = 0$$ QED :-)