A quick recap of both the examples that made it to the blackboard, and
some I had in mind if time permitted.
There are a couple of approaches to questions of limit and continuity. All
rely on a clear understanding of what the terms mean. Say we have a function
f of two variables x, y into the real numbers R and we’re interested in whether
or not lim(x,y)→(x0 ,y0 ) f (x, y) = L. To prove that the answer is ‘yes’, we have
to show that for any interval I about L, , there’s a disk about (x0 , y0 ) so that
the whole thing, with the possible exception of (x0 , y0 ) itself, is mapped by f
into that interval I. To prove that the answer is ‘no’, we have to show that
there’s an interval I about L so that for any disk about (x0 , y0 ), (SOMETHING
HAPPENS.) Try and remember what that ‘something’ is. The answer is at the
bottom of this post.
In all our examples, (x0 , y0 ) is (0, 0), and L = 0.
First example: f (x, y) = xy/(x2 + y 2 ). There is no limit. The short answer
to ‘is the limit 0?’ is ‘no’. The more thorough answer is ‘no’, because there
isn’t any number L that can be called the limit. And why not? Because for
the interval (−1/4, 1/4) about 0, inside any disk about (0, 0) there are nonzero
points (c, c) inside that disk and f (c, c) = c2 /(2c2 ) = 1/2 which is outside
(−1/4, 1/4).
Second example: Now f (x, y) = x2 y/(x2 + y 2 ). This time around, when we
try x = y = c, we find that xy 2 /(x2 + y 2 ) = c/2. Whatever interval we use,
even if it’s smaller than (−1/4, 1/4), if we choose a small enough radius for our
disk, points (c, c) inside that disk will map to a c/2 that lies inside our target
interval. The attempt to prove that the answer is ‘no’ by way of the path y = x
toward (x0 , y0 ) = (0, 0) fails. But what if there would be another path that gave
a different answer? We cannot examine every possible path, one by one.
New idea: use inequalities. Here are some basic inequalities that tend to
come in handy. Each is valid for any choice of real numbers a and b:
√
|a| ≤ a2 + b2
1
|ab| ≤ (a2 + b2 )
2
How might these be applied to x2 y/(x2 + y 2 )? There are some silly things we
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might do here, such as argue that x2 y/(x2 + y 2 ) ≥ x2 y/((x
√ + y ) + y ), using
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the inequality to replace each x in the denominator with x + y . Steps that
make things messier and less symmetric are likely steps in the wrong√direction.
Instead, how about replacing every x and y in the numerator with x2 + y 2 ?
That makes the expression larger. How is this good? We want it to be small!
But the point is, we want to prove that it was originally small. If we replace
it with something bigger, and prove that the replacement is small, then the
original was also small.
Our step yields this information:
x2 y (x2 + y 2 )3/2
≤
|f (x, y)| = 2
= (x2 + y 2 )1/2 .
x + y2 x2 + y 2
1
Now we could again think about target intervals and disks. If the radius of
the target interval is ϵ, and we choose a disk of radius ϵ/2 about 0 and take
some
nonzero (x, y) from inside that disk, then by the meaning of ‘radius’,
√
x2 + y 2 ≤ ϵ/2 < ϵ. So the whole disk, apart from an undefined output from
(0, 0), gets mapped into our interval I. Proved.
The less formal but perfectly acceptable way to close out the proof is simply
to observe that the limit of (x2 + y 2 )1/2 as (x, y) → (0, 0) is 0. Either way, we’re
done and the limit exists; it’s 0.
Next example: f (x, y) = |x + y| log |x − y|. Here, there’s a pitfall. The line
y = x lies above the graph of y = log x. This means that for u > 0, log u < u. So
we might apply this and arrive at f (x, y) ≤ |x + y||x − y| = |x2 − y 2 | ≤ |x2 + y 2 |.
And that’s perfectly correct. It is, unfortunately, not the kind of inequality we
need, because we need to prove that f (x, y) is close to L when (x, y) is near
(x0 , y0 ). Here, that means we need to prove that f (x, y) is close to 0. But all we
have is that f (x, y) lies somewhere below |x2 − y 2 |. Perhaps |x2 − y 2 | = 1/100,
and f (x, y) = −3? After all, −3 < 1/100! So we don’t get a proof that the limit
is zero.
The failure of the attempt at a proof is not a washout, though. We can
salvage an insight from it: perhaps our f is too seriously negative at points near
the origin? And indeed, that’s just what happens. When y = x, f isn’t even
defined. And if we rule out those cases, saying that this line is the boundary of
the domain of f , we still have points such as (1/10, 1/10 + 1/1000000000000)
that lie inside the domain but for which log |x − y| is something akin to −25.
The limit does not exist because there are situations like this as near to (0, 0)
as you like.
Our last example is f (x, y) = |x + y| log(x2 + xy + y 2 ). The domain of this f
is the whole plane apart from (0, 0), because the input to log here, x2 +xy+y 2 , is
sure to be positive when either x or y is nonzero. Because why? Because |xy| ≤
(1/2)(x2 + y 2 ), so in particular, −xy ≤ (1/2)(x2 + y 2 ). So xy ≥ −(1/2)(x2 + y 2 )
and x2 + xy + y 2 ≥ (1/2)(x2 + y 2 ). (When y = −x, the inequality becomes
exact.) Now why are we focused on proving that x2 + xy + y 2 is greater or equal
to something? Because when we take the log, the danger is that it would be
very negative.
To continue, we replace | log(x2 + xy + y 2 )| with | log((1/2)(x2 + y 2 ))| and
|x + y| with 2x2 + 2y 2 , the second with two applications of one of our featured
basic inequalities, and we know that our new expression is an upper bound for
the absolute value of the original. But observe: our new expression depends not
on the particular choice of x and y, but just on the radius r associated with
x and y. All we need do now is show that limr→0 2r2 | log(r2 /2)| = 0. That’s
a question for first semester calculus and L’Hopital’s rule works like a charm.
The limit is indeed 0 and we’re done.
What was that ‘something happens’ from the question set at the top of the
post? For any disk about (x0 , y0 ), some part of the image under f lies OUTSIDE
the target interval I.
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