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Epsilon delta proof of a two-variable limit using inequalities

LeibnizIsBetter

MHB Donor
Aug 28, 2013
3
I seem to be having trouble with multivariable epsilon-delta limit proofs. I don't have a very good intuition for how \(\displaystyle \epsilon\) relates to \(\displaystyle \delta\).

For example:

\(\displaystyle Prove \lim_{(x,y) \to (0,0)}\frac{2xy^2}{x^2+y^2} = 0\)

There are probably many ways to do this, but my teacher does it a certain way and I would like to learn his way first (although I am also interested in other techniques–perhaps using the polar coordinate system?).

He "massages" \(\displaystyle \left|f(x,y) - L\right| < \epsilon\) into a form compatible with \(\displaystyle \sqrt{(x-a)^2 + (y-b)^2} < \delta\) by using a series of inequalities.

\(\displaystyle
\left|\frac{2xy^2}{x^2 + y^2} - 0\right|
= \left|\frac{2\sqrt{x^2}y^2}{x^2 + y^2}\right|
\leq \left|\frac{2\sqrt{x^2 + y^2}y^2}{x^2 + y^2}\right|
= \left|\frac{2y^2}{\sqrt{x^2 + y^2}}\right|
\leq \left|\frac{2(y^2 + x^2)}{\sqrt{x^2 + y^2}}\right|
= 2\sqrt{x^2 + y^2}
\)

Step by step, I get this. But I'm not getting the bigger picture.

Could someone please explain to me how to use the above inequalities to prove the limit equals zero?

Thanks in advance. Any help is appreciated.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
If you're not really understanding the [tex]\displaystyle \begin{align*} \epsilon - \delta \end{align*}[/tex] definitions of a limit, it might help with a metaphor.

When I do [tex]\displaystyle \begin{align*} \epsilon - \delta \end{align*}[/tex] proofs, I think of myself pulling pizzas out of an oven (I used to work in a pizza shop). Think of there being an "ideal" level of cooking for your pizza. Obviously, it is not going to be possible to get this "ideal" amount of cooking for every pizza (or possibly even any pizza), but there is a certain "tolerance" you can have for over-cooking or under-cooking before you consider it raw or burnt. As long as you are reasonably close to the right amount of time needed, then your level of cooking will be considered acceptable. Then as you gain more experience, you should be able to get closer and closer to keeping the pizzas in the oven for the ideal amount of time, thereby making your pizzas closer and closer to the ideal level of cooking, which means you would expect that your tolerance would decrease as you'd be getting used to your pizzas being cooked properly.

So if we were to call the amount of time in the oven [tex]\displaystyle \begin{align*} x \end{align*}[/tex], then the level of cooking is some function of x [tex]\displaystyle \begin{align*} f(x) \end{align*}[/tex]. We said there is an ideal level of cooking, we could call that [tex]\displaystyle \begin{align*} L \end{align*}[/tex], which means there is a point in time [tex]\displaystyle \begin{align*} x = c \end{align*}[/tex] which gives this ideal level of cooking. Remember we said that as long as we have kept the pizzas in the oven for an amount of time reasonably close to [tex]\displaystyle \begin{align*} c \end{align*}[/tex], say [tex]\displaystyle \begin{align*} \delta \end{align*}[/tex] units of time away from it, then our level of cooking would be considered acceptable, or within some tolerance which we could call [tex]\displaystyle \begin{align*} \epsilon \end{align*}[/tex]. So we need to show that [tex]\displaystyle \begin{align*} \delta \end{align*}[/tex] and [tex]\displaystyle \begin{align*} \epsilon \end{align*}[/tex] are related, so that you are guaranteed that as you get experience and keep your pizzas in the oven closer to the right amount of time ( i.e. [tex]\displaystyle \begin{align*} \delta \end{align*}[/tex] gets smaller) then so will your tolerance [tex]\displaystyle \begin{align*} \epsilon \end{align*}[/tex] get smaller and closer to the ideal level of cooking.

Do you see now what it means to show [tex]\displaystyle \begin{align*} 0 < |x - c| < \delta \implies |f(x) - L | < \epsilon \end{align*}[/tex]? It means if you have set a tolerance around your ideal limiting value, then as long as you are reasonably close to [tex]\displaystyle \begin{align*} x = c \end{align*}[/tex], then you are guaranteed that your function value is within your tolerance away from the limiting value, and by showing the relationship between [tex]\displaystyle \begin{align*} \delta \end{align*}[/tex] and [tex]\displaystyle \begin{align*} \epsilon \end{align*}[/tex], you are guaranteed that as your [tex]\displaystyle \begin{align*} \delta \end{align*}[/tex] gets smaller and you close in on [tex]\displaystyle \begin{align*} x = c \end{align*}[/tex], then your tolerance will get smaller and your [tex]\displaystyle \begin{align*} f(x) \end{align*}[/tex] will close in on [tex]\displaystyle \begin{align*} L \end{align*}[/tex].


Mind you, this is only for a one-variable situation. But the same intuition applies for two-variable. The difference is that you are trying to show that no matter what point you start from, and what path you take to close in on the point, that being a certain distance close enough to the point you are approaching is enough to ensure that you are going to be within your tolerance. Surely you remember from high school that the distance between two points [tex]\displaystyle \begin{align*} (x,y) \end{align*}[/tex] and [tex]\displaystyle \begin{align*} (h,k) \end{align*}[/tex] can be found using Pythagoras, i.e. [tex]\displaystyle \begin{align*} \sqrt{ \left( x - h \right) ^2 + \left( y - k \right) ^2} \end{align*}[/tex]. So to summarise what you are trying to do, you want to show that as long as your distance between the point where you are [tex]\displaystyle \begin{align*} (x, y) \end{align*}[/tex] is close enough to the point you are approaching [tex]\displaystyle \begin{align*} (h, k) \end{align*}[/tex], i.e. [tex]\displaystyle \begin{align*} \delta \end{align*}[/tex] units, then you are guaranteed that your function will always be within the tolerance [tex]\displaystyle \begin{align*} \epsilon \end{align*}[/tex] from your ideal/limiting value[tex]\displaystyle \begin{align*} L \end{align*}[/tex].

So to prove that [tex]\displaystyle \begin{align*} \lim_{(x, y) \to (h, k)} f(x,y) = L \end{align*}[/tex] you have to show that [tex]\displaystyle \begin{align*} 0 < \sqrt{ \left( x - h \right) ^2 + \left( y- k \right) ^2 } < \delta \implies \left| f(x, y) - L \right| < \epsilon \end{align*}[/tex].

Hope that helped...
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,706
I seem to be having trouble with multivariable epsilon-delta limit proofs. I don't have a very good intuition for how \(\displaystyle \epsilon\) relates to \(\displaystyle \delta\).

For example:

\(\displaystyle Prove \lim_{(x,y) \to (0,0)}\frac{2xy^2}{x^2+y^2} = 0\)

There are probably many ways to do this, but my teacher does it a certain way and I would like to learn his way first (although I am also interested in other techniques–perhaps using the polar coordinate system?).

He "massages" \(\displaystyle \left|f(x,y) - L\right| < \epsilon\) into a form compatible with \(\displaystyle \sqrt{(x-a)^2 + (y-b)^2} < \delta\) by using a series of inequalities.

\(\displaystyle
\left|\frac{2xy^2}{x^2 + y^2} - 0\right|
= \left|\frac{2\sqrt{x^2}y^2}{x^2 + y^2}\right|
\leq \left|\frac{2\sqrt{x^2 + y^2}y^2}{x^2 + y^2}\right|
= \left|\frac{2y^2}{\sqrt{x^2 + y^2}}\right|
\leq \left|\frac{2(y^2 + x^2)}{\sqrt{x^2 + y^2}}\right|
= 2\sqrt{x^2 + y^2}
\)

Step by step, I get this. But I'm not getting the bigger picture.

Could someone please explain to me how to use the above inequalities to prove the limit equals zero?

Thanks in advance. Any help is appreciated.
The most simple-minded way of thinking about the fraction \(\displaystyle \frac{2xy^2}{x^2 + y^2}\) is that the numerator is a cubic expression in $x$ and $y$, whereas the denominator is quadratic. So when $x$ and $y$ are small, the numerator has "three orders of smallness" but the denominator has only two. So the fraction as a whole has one "order of smallness" and therefore tends to $0$ as $(x,y)\to(0,0).$ Of course, that does not qualify as a mathematical argument, but you can make it rigorous either by your teacher's "massaging" method, or (in the case of this example) by using the fact that $|2xy|\leqslant x^2+y^2$. (That follows from the fact that $(|x|-|y|)^2\geqslant0$.) You can then see that \(\displaystyle \left|\frac{2xy^2}{x^2 + y^2}\right| = \left|\frac{(2xy)y}{x^2 + y^2}\right| \leqslant |y| \leqslant \sqrt{x^2+y^2}\), and now you can feed in the epsilons and deltas to complete the proof that \(\displaystyle \lim_{(x,y) \to (0,0)}\frac{2xy^2}{x^2+y^2} = 0\).
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
For functions of two variables, like this, with (x, y) going to (0, 0), you might find it easier to switch to polar coordinates. That way the distance from (0, 0) is measured by the single variable, r.

Here, the denominator is [tex]x^2+ y^2[/tex] which is simply "[tex]r^2[/tex]". The numerator is [tex]2xy^2= 2(r cos(\theta))(r^2 sin^2(\theta))[/tex] so the fraction is
[tex]\frac{2r^3sin^2(\theta)cos(\theta)}{r^2}= 2rsin^2(\theta)cos(\theta)[/tex] and, as r goes to 0, that goes to 0 no matter what [tex]\theta[/tex] is. As for "[tex]\epsilon[/tex], [tex]\delta[/tex]", it is sufficient to note that [tex]2sin^2(\theta)cos(\theta)\le 2[/tex], so that [tex]|2r sin^2(\theta)cos(\theta)|\le 2 r[/tex]. Given any [tex]\epsilon> 0[/tex], we may choose [tex]\delta\le \frac{1}{2}\epsilon[/tex].