Domains, level sets and multivariable limits

Try

Consider the height z=√(4−x²−y²). Before plotting it, determine where a real height exists. The radicand is nonnegative exactly on the closed disk x²+y²≤4. At the center the height is 2, and on the boundary it is zero. This graph is an upper hemisphere. Its domain is a set of input pairs; its graph is a set of triples. Confusing these objects can make a perfectly valid boundary point disappear from a calculation.

Predict

Predict the shape of the level set z=1. Squaring gives x²+y²=3, a circle inside the domain. In contrast, level sets of f(x,y)=x²+4y² are ellipses, and those of f(x,y)=x²−y² are hyperbolas except at the zero level, where two lines cross. Draw several levels with labels. Equal changes in the level label do not necessarily correspond to equal distances between curves; crowded contours often indicate rapid change.

Experiment

Inspect g(x,y)=x²/(x²+y²) away from the origin. On the horizontal axis its value is 1; on the vertical axis its value is 0. Both paths approach the same point, so there cannot be a single limiting value there. Changing the function’s assigned value at the origin cannot repair that mismatch.

Approach	Substitution	Value away from origin
y=0	x²/x²	1
x=0	0/y²	0
y=x	x²/(2x²)	1/2

Use the table to explain why an apparent smooth trace along one selected path is insufficient evidence for a multivariable limit.

Understand

A limit L at a point p requires f(x) to approach L whenever the distance |x−p| tends to zero through domain points other than p. In epsilon-delta language, every desired output tolerance ε>0 must have a distance tolerance δ>0 that works for all such points within δ. The tolerance cannot depend on which approach path was chosen.

To prove a limit, seek a bound involving distance. For h(x,y)=x²y²/(x²+y²), the inequality x²y²≤(x²+y²)²/4 follows from (x²−y²)²≥0. Thus 0≤h≤(x²+y²)/4, which tends to zero. Defining h(0,0)=0 makes it continuous. This is an all-directions argument, unlike checking several examples.

Apply

Repair the claim “every straight line gives limit zero, therefore the full limit is zero.” Consider k(x,y)=x²y/(x⁴+y²). Along y=mx with fixed nonzero m, simplification gives mx/(x²+m²), which tends to zero; along y=0 the value is zero. Along the curved path y=x², however, the value is 1/2. Every straight-line test can agree while a curved path reveals failure.

For a quotient of continuous functions, continuity holds wherever the denominator is nonzero. At excluded points, that rule gives no conclusion. A limit may still exist and permit an extension, or it may fail as above. Determine which by a bound or a counterexample.

Continue

As optional depth, compare boundary-relative continuity with continuity on an open neighborhood. The hemisphere height is continuous on its closed-disk domain even at boundary points, because inputs are restricted to that domain. Later derivative theorems often assume open neighborhoods, so the same point needs additional attention there.

Record one nonexistence proof with two paths and one existence proof with a distance bound. These are different logical tasks: counterexamples disprove a universal limit claim, whereas selected successful paths cannot establish it. Continue to partial derivatives, retaining the habit of asking whether an argument controls every small displacement or only a preferred slice.