Homework Help: 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.