Prove or disprove each of the follow function has limits $x \to a$ by the definition $\lim_{(x, y) \to (0, 0)} \frac{x^2y}{x^2 + y^2}$

Question

Prove or disprove each of the follow function has limits $x \to a$ by the definition

$\lim_{(x, y) \to (0, 0)} \frac{x^2y}{x^2 + y^2}$

Let $y = x^2$

$\frac{x^2 y}{x^2 + y^2} = \frac{x^4}{2x^4} = \frac{1}{2}$

If we let $x = 0$, then

$\lim_{(x, y) \to (0,0)} \frac{x^2y}{x^2+y^2} = \frac{0^2 \cdot 0}{0^2 + 0^2} = 0$

Therefore $\lim_{(x, y) \to (0, 0)} \frac{x^2y}{x^2 + y^2} = 0 \neq 1/2$

Therefore the limit does not exist because two different values.

Would this be correct?

You have asked a similar question at here. These two limits are identical since $x,y$ are symmetric here, so the answer should be the same as well. Then why did you conclude differently for these two? — xbh, Oct 18 '18 at 01:55

score 0 · Accepted Answer · answered Oct 18 '18 at 01:49

0

Note that $$ \left|\frac{x^{2}y}{x^{2}+y^{2}}\right|=\frac{x^{2}|y|}{x^{2}+y^{2}}=\frac{x^{2}\sqrt{y^{2}}}{x^{2}+y^{2}}\leq\frac{(x^{2}+y^{2})\sqrt{y^{2}}}{x^{2}+y^{2}}=\sqrt{y^{2}}\leq\sqrt{x^{2}+y^{2}}=\|(x,y)\|\to 0. $$

answered Oct 18 '18 at 01:49

ervx

12,208

So, the limit does exist ? – Tree Garen Oct 18 '18 at 08:59
Yes, it's zero. – ervx Oct 18 '18 at 12:34

score 0 · Answer 2 · answered Oct 18 '18 at 01:52

Note that if $y=x^2$, we have

$$\lim_{x \to 0}\frac{x^2 \cdot x^2}{x^2+x^4}=\lim_{x \to 0}\frac{x^2 }{1+x^2}=0$$

If we let $x=0$, we havae

$$\lim_{y \to 0}\frac{0^2 \cdot y}{0^2+y^2}=\lim_{y \to 0}0=0$$

In fact, the limit is $0$ since

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

Prove or disprove each of the follow function has limits $x \to a$ by the definition $\lim_{(x, y) \to (0, 0)} \frac{x^2y}{x^2 + y^2}$

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