# Delta/epsilon two variable limit

1. Apr 7, 2014

### iRaid

1. The problem statement, all variables and given/known data
In calculus 1, honestly, I never understood this as it just doesn't make sense. Now in multivariable calculus, it still doesn't make sense. After all these engineering, math, and science classes, I still cannot figure these problems out, and it pisses me off!! I would appreciate it if someone could explain this example to me...

Evaluate
$$\lim_{(x,y)\rightarrow (0,0)} \frac{5x^2y}{x^2+y^2}$$

2. Relevant equations

3. The attempt at a solution
This is an example problem, the book does not explain this well at all. The answer they get is: You can choose $\delta = \varepsilon /5$ and conclude that the limit = 0.

I don't even see how they just (seemingly) magically conclude that.
Any help is appreciated.

2. Apr 7, 2014

### LCKurtz

Try expressing the variables in polar coordinates. $(x,y)\to (0,0)$ is the same as $r\to 0$.

3. Apr 7, 2014

### az_lender

The definition of a limit would imply that the limit will be 0 if, for every small positive ε, there exists some δ such that |5xy2/(x2 + y2)| < ε whenever the distance from (x,y) to (0,0) is less than δ. We're can always ignore LARGE values of ε, so just assume ε < 1 (you'll see why).

If x2+y2 < δ2, you know that x2 <= (x2+y2) < δ2 < δ; the last inequality depends on δ < ε < 1. So if δ = ε/5, then 5x2 < 5δ = ε. And you know that y/(x2+y2) is sin(θ), whose absolute value is <= 1. That proves the absolute value of the entire fraction is smaller than ε.

4. Apr 7, 2014

### LCKurtz

No, $\frac y {\sqrt{x^2+y^2}}=\sin\theta$.

5. Apr 7, 2014

### iRaid

So I can just change these functions into polar coordinates and solve that way?

Is this the best way to approach these problems?

6. Apr 7, 2014

### LCKurtz

There is no best way. What you do depends on the nature of the problem. In this problem the degree of the numerator is 3 and the denominator is 2, plus the $x^2+y^2$ in the denominator suggest trying polar coordinates. Try it.

7. Apr 7, 2014

### Zondrina

$$∀ε>0,∃δ>0 \space | \space 0<||(x,y)-(0,0)||<δ⇒|\frac{5x^2 y}{x^2+y^2}-0|<ε$$

Note that $||(x,y)-(0,0)||<δ⇒|x|<δ∧ |y|<δ$.

By the triangle inequality:

$|\frac{5x^2 y}{x^2+y^2}-0|= \frac{5|x|^2 |y|}{|x^2+y^2|} ≤ \frac{5|x|^2 |y|}{|x|^2+|y|^2 } < \frac{5δ^3}{2δ^2} < \frac{5δ^3}{δ^2} = 5δ$

So we have:

$5δ ≤ ε ⇒ δ ≤ \frac{ε}{5}$

Therefore, by choosing $δ ≤ \frac{ε}{5}, |\frac{5|x|^2 |y|}{|x^2+y^2|}-0|<ε$. Although you should try the polar co-ordinates as well as it is often a useful trick. Try proving the chosen $\delta$ does indeed satisfy the definition.

Last edited: Apr 7, 2014
8. Apr 7, 2014

### jbunniii

Perhaps simpler:

If $x = 0$ and $y \neq 0$, then we have
$$\frac{5x^2y}{x^2+y^2} = 0$$

If $x \neq 0$ we may write
$$\frac{5x^2y}{x^2+y^2} = \frac{5y}{1 + y^2/x^2}$$
As $y \rightarrow 0$, the numerator goes to zero. The denominator is at least $1$ for all $x,y$. What can you conclude?

9. Apr 8, 2014

### LCKurtz

If we are going to explore alternatives, you can also use $|xy|\le \frac{x^2+y^2} 2$ which comes from $(|x|-|y|)^2\ge 0$. Then$$\left|\frac{5x^2y}{x^2+y^2}\right|\le 5 |x|\left|\frac{xy}{x^2+y^2}\right|\le\frac 5 2 |x|$$which leads to a solution too. So @iraid, I guess you have seen more than one way to skin this particular cat. Still, I think polar coordinates is the simplest.