# Math Help - Epsilon-Delta for Multivariable functions

1. ## Epsilon-Delta for Multivariable functions

Ok, epsilon-delta was the only thing that gave me trouble in calculus in high school and now in Calc III in college, it has only gotten worse.

I have looked at the other examples on here for multivariable limit proofs, and most of them either are slightly trivial because the point the limit is being evaluated at is (0,0), or I am unable to follow the logic. It seems like, many times, people just pull a value for delta, based on epsilon, out of thin air.

My problem is:

lim (x,y) -> (5,2) of 4xy = 40

I have the implication set up, 0< [(x-5)^2 + (y-2)^2]^(1/2) < delta
implies
|(4xy)-40| < epsilon

but I don't know how to manipulate that properly.

It's particularly frustrating because this topic isn't even in our textbook- our professor just foisted it on us with little to no preface.

2. Originally Posted by WebsterCalculus
Ok, epsilon-delta was the only thing that gave me trouble in calculus in high school and now in Calc III in college, it has only gotten worse.

I have looked at the other examples on here for multivariable limit proofs, and most of them either are slightly trivial because the point the limit is being evaluated at is (0,0), or I am unable to follow the logic. It seems like, many times, people just pull a value for delta, based on epsilon, out of thin air.

My problem is:

lim (x,y) -> (5,2) of 4xy = 40

I have the implication set up, 0< [(x-5)^2 + (y-2)^2]^(1/2) < delta
implies
|(4xy)-40| < epsilon

but I don't know how to manipulate that properly.

It's particularly frustrating because this topic isn't even in our textbook- our professor just foisted it on us with little to no preface.

I kept stepping though the pages till i found a thread un-answered.

you're on the right path. All u have to do, at least it's what i did, is parametrize your relation between delta and (x,y).

that is, consider the circle with radius $\delta$ about (5,2).

Then for any point (x,y) on the circle,
$(x-5)^2+(y-2)^2 = \delta^2$
Now if we parametrize $x$ and $y$ to be functions of another variable $\theta$, something i'm sure you've seen by now if you're in Calc III, then
$x = 5+\delta\cos\theta < 5 + \delta$
$y = 2+\delta\sin\theta < 2 +\delta$
because $\sin(x),\cos(x)<1 \forall x$

Then

$4xy-40<4(5+\delta)(2+\delta)-40 = 40+28\delta+4\delta^2 - 40$ = $4\delta^2+28\delta$

Now pick an $\epsilon$.
Set $\delta$ equal to the $\min({r_1,r_2})$ where $r_1,r_2$ are the roots of the equation:
$\delta^2+7\delta-\frac{\epsilon}{4}$