Limit definition (2 variables)

Question

I have the following function:

$$ f(x,y) = \begin{cases} \dfrac{xy^3}{{x^2+y^2}} & (x,y)\neq(0,0) \\ 0 & (x,y)=(0,0) \\ \end{cases} $$

How can I demonstrate the existence of such a limit using epsilon and delta?

Answer 1

Presumably, the limit you want to check is $$\lim_{(x,y)\to(0,0)} f(x) = \lim_{(x,y)\to(0,0)} \frac{xy^3}{x^2+y^2}$$

There are some useful inequalities that help with questions like this. One comes from noting the following. Since $(x+y)^2\ge 0$, and $(x-y)^2\ge 0$, we can expand and rearrange these to conclude that $|xy|\le x^2+y^2$, for all $x$ and $y$. (In fact, you get that $2|xy|\le x^2+y^2$, but we can use the weaker version.)

Now, you can rewrite the expression in question as $\frac{xy}{x^2+y^2}\cdot y^2$. By the inequality derived above, the first fraction is always $\le 1$ in absolute value. Then, we know that $y^2\to 0$ as $(x,y)\to (0,0)$. To set up your $\epsilon-\delta$ argument, you just need to know that $y^2<\epsilon$, which will happen if $y<\sqrt{\epsilon}$.

Setting $\delta=\sqrt{\epsilon}$, we can now argue that $|f(x,y)|<\epsilon$ under various circumstances, such as $\max\{|x|,|y|\}<\delta$, or $\sqrt{x^2+y^2}<\delta$, or $|x|+|y|<\delta$, any of which puts us in a small neighborhood of the origin.

Answer 2

Let $\varepsilon > 0$.

Choose $\delta = \sqrt{\varepsilon}$.

For $\|(x,y)\|_2 < \delta$ we have:

$$\left|\frac{xy^3}{x^2 + y^2}\right| = \frac{|x||y|^3}{\|(x,y)\|_2^2} = \underbrace{\frac{|x|}{\|(x,y)\|_2}}_{\le 1} \cdot \underbrace{\frac{|y|}{\|(x,y)\|_2}}_{\le 1} \cdot\underbrace{|y|^2}_{<\delta^2} < \delta^2 = \varepsilon$$

since $\max\{|x|,|y|\} \le \|(x,y)\|_2$.

Thus $\displaystyle\lim_{(x,y)\to(0,0)} \frac{xy^3}{x^2 + y^2} = 0$ so $f$ is continuous at $(0,0)$.

Answer 3

$ |f(x,y) - f(0,0)| = $

$|\dfrac{xy^3}{x^2+y^2}| \le \dfrac{|xy|(x^2+y^2)}{(x^2+y^2)} =$

$|xy| \lt (x^2+y^2)$.

Let $\epsilon >0$ be given.

Choose: $\delta= \epsilon^{1/2}$.

Then $|x^2+y^2|^{1/2} \lt \delta$ implies

$|f(x,y)-f(0,0)| \lt $

$(x^2+y^2)\lt \delta^2 =\epsilon.$

Note: $x^2+y^2 \ge 2|xy| \ge |xy|$.