Hi Ashley,
We must show that if we make (x, y) sufficiently close to (1, 1), then the function value f(x, y) = 4x + (2*(y^2)) will be as close as we like to 6.
More formally, we must show that for every epsilon > 0, there exists a delta > 0 such that if
0 < dist((x, y), (1, 1)) < delta, then f(x, y) will be between 6 – epsilon and 6 + epsilon.
Before we launch into a formal proof, let's play around a little bit with how the function f(x, y) behaves when we move x and y a little bit away from (1, 1). Observe that if I change x by a bit, 4x changes by 4 times as much. (E.g., when x goes from 1 to 1.01, 4x goes from 4 to 4.04.) What about 2y^2? Basic calculus tells us that its derivative is 4y, which at y = 1 is equal to 4. So if we change y by a little bit, 2y^2 also changes by about 4 times as much, or maybe a tiny bit more. (E.g., when y goes from 1 to 1.01, 2y^2 goes from 2 to 2.0402 which is just a tad more than 2.04.)
If (x, y) is within some (small) distance delta of (1, 1), then certainly x will be at most delta from 1, and y will be at most delta from 1. Thus 4x will be at most 4*delta from 4. As we saw above, 2y^2 could be a bit more than 4*delta from 2. So, at worst, 4x + 2y^2 will be a bit more than 8*delta from 6. Hopefully we can show that, say, 4x + 2y^2 will be less than 10*delta from 6. If so, then we can let delta = epsilon/10, and we'll have that 4x + 2y^2 is less than epsilon from 6, which is what we want.
OK, now here goes with the formal argument. Let delta = min(epsilon/10, 1), that is, the lesser of epsilon/10 of 1. (Why am I doing this minimum business? Say epsilon = 1000. I don't want to allow delta to be 100, because then y could be up to 101, and 2*101^2 = 20,402, which is WAY more than 1000 away from 6. Because of the square term, big delta can make f(x, y) REALLY far off from 6. So if epsilon is really big, let's just cap delta off at 1.)
As we said above, since the distance from (x, y) to (1, 1) is less than delta, x and y must be less than delta from 1; that is, |x – 1| < delta and |y – 1| < delta. The first inequality immediately implies that |4x – 4| < 4*delta, i.e., the distance from 4x to 4 is less than 4*delta.
I claim that |2y^2 – 2|, the distance from 2y^2 to 2, is less than 2*delta^2 + 4*delta. We know that y lies in the open interval (1 – delta, 1 + delta). Since delta <= 1, the left endpoint of the interval 1 – delta >= 0. Thus, y must be positive. When y is positive, 2y^2 is increasing (its derivative is positive).
So we have to look at how far above 2 the value 2y^2 is toward the right side of our open interval, and how far below 2 the value 2y^2 is toward the left side of our open interval, and then see which of these distances is greater. At y = 1 + delta, we have 2y^2 = 2*delta^2 + 4*delta + 2, which is 2*delta^2 + 4*delta above 2. At y = 1 – delta, we have 2y^2 = 2*delta^2 – 4*delta + 2. The quantity 2*delta^2 – 4*delta is negative, so this is -(2*delta^2 – 4*delta) = -2*delta^2 + 4*delta below 2. Note that -2*delta^2 + 4*delta is less than 2*delta^2 + 4*delta. Thus, on our interval, 2y^2 will be at most 2*delta^2 + 4*delta away from 2; that is, |2y^2 – 2| < 2*delta^2 + 4*delta. Phew!
Now we're almost there. Since 4x is less than 4*delta away from 4, and 2y^2 is less than 2*delta^2 + 4*delta away from 2:
The distance from 4x + 2y^2 to 6 will be less than 4*delta + (2*delta^2 + 4*delta) = 2*delta^2 + 8*delta.
Since delta <= 1, 2*delta^2 <= 2*delta, so:
The distance from 4x + 2y^2 to 6 is less than 2*delta + 8*delta = 10*delta.
Finally, delta <= epsilon/10, so:
The distance from 4x + 2y^2 to 6 is less than 10*delta <= 10*(epsilon/10) = epsilon.
QED!
I hope this makes sense. By the way, note that although delta = min(epsilon/10, 1) worked well for us, we could have chosen a different, smaller delta, e.g. delta = min(epsilon/100, 1), and that would have worked fine, too. Making delta smaller in this way only makes 4x + 2y^2 even closer to 6, so it wouldn't hurt. You don't have to find an optimally-sized delta; if it helps, choose delta to be really, really tiny, much smaller than is necessary. You're allowed to make delta as small a positive number as you like, as long as it gets you within epsilon of the desired limit.
Best wishes,
Kiran
