Limits and continuity in several variables

Setting up the definition. For a function $f(x,y)$ defined near $(a,b)$ (though not necessarily at $(a,b)$), the limit statement generalizes the familiar $\varepsilon$-$\delta$ idea from one variable, except distance is now measured in the plane rather than along a line.

where: $\varepsilon$ is the tolerance on the output, $\delta$ is the corresponding tolerance on how close $(x,y)$ must sit to $(a,b)$, and $\sqrt{(x-a)^2+(y-b)^2}$ is ordinary Euclidean distance in the plane, the two-variable stand-in for $|x-a|$.

Why "every path" matters. In $\mathbb{R}^1$ there are exactly two directions of approach, so two one-sided limits settle the question. In $\mathbb{R}^2$ there is a whole continuum of directions and even curved routes of approach, and the definition above only tests a small disc shrinking around $(a,b)$ — it says nothing about which route you take through that disc. That silence is precisely the point: the definition demands agreement along every route simultaneously. This gives a one-directional test that is used constantly in practice — if two specific paths through $(a,b)$ produce two different limiting values, the two-variable limit is immediately certified not to exist, since a genuine limit would have forced both paths to agree. The reverse implication does not hold: agreement along finitely many paths, however many you check, never proves existence, because infinitely many untested paths remain.

Continuity. Once the limit machinery is in place, continuity is a short add-on: a function is continuous at $(a,b)$ when the limit exists there and equals the function's actual value.

where: the left side requires the limit to exist at all (agreement along every path), and the equality then pins that common value to $f(a,b)$ itself, ruling out removable-looking gaps. As in single-variable calculus, sums, differences, products, quotients (away from zero denominators), and compositions of continuous functions stay continuous, so polynomials in $x,y$ are continuous everywhere and rational functions are continuous wherever the denominator does not vanish — you rarely need to grind through $\varepsilon$-$\delta$ by hand once you know this.

Show that $\displaystyle \lim_{(x,y)\to(0,0)} \frac{x^2y}{x^2+y^2} = 0$, and use this to establish that $f(x,y) = \dfrac{x^2y}{x^2+y^2}$, with $f(0,0)=0$, is continuous at the origin.

Step 1. Substitute $x=r\cos\theta,\ y=r\sin\theta$, so $(x,y)\to(0,0)$ becomes $r\to0^+$ with $\theta$ arbitrary:

Step 2. Bound the angular factor. Since $|\cos^2\theta\sin\theta| \le 1$ for every $\theta$ (each factor is bounded by $1$ in magnitude), we get $|f(x,y)| = r\,|\cos^2\theta\sin\theta| \le r$.

Step 3. Squeeze. As $r\to0^+$ the bound $r\to0$ independently of $\theta$, so by the squeeze theorem $f(x,y)\to0$ along every path simultaneously; the limit exists and equals $0$.

Step 4. Check continuity. We are told $f(0,0)=0$, and Step 3 gave $\lim_{(x,y)\to(0,0)} f(x,y) = 0 = f(0,0)$, so $f$ is continuous at the origin. Away from the origin $f$ is a quotient of polynomials with nonzero denominator, hence continuous there too, so $f$ is continuous on all of $\mathbb{R}^2$.

Competitive stretch

Does $\displaystyle \lim_{(x,y)\to(0,0)} \frac{x^3y}{x^6+y^2}$ exist?

Step 1. Try every straight line $y=mx$ first. Then $f(x,mx) = \dfrac{x^3 \cdot mx}{x^6+m^2x^2} = \dfrac{mx^4}{x^2(x^4+m^2)} = \dfrac{mx^2}{x^4+m^2} \to 0$ as $x\to0$, for every fixed slope $m$ (including $m=0$). So all straight-line paths, without exception, agree on the value $0$.

Step 2. That agreement is the trap the traps section below warns about — try the curved path $y=x^3$ instead. Then $f(x,x^3) = \dfrac{x^3\cdot x^3}{x^6+x^6} = \dfrac{x^6}{2x^6} = \dfrac12$ for all $x\neq0$, so $f\to\dfrac12$ along this path.

Step 3. Compare. Straight lines give $0$; the curve $y=x^3$ gives $\dfrac12$ — two different values. The limit does not exist, even though it looked settled after checking infinitely many straight lines.

• Checking only straight-line paths $y=mx$ and declaring the limit exists. Agreement along all lines is necessary but never sufficient, as the competitive stretch example shows — a matched-degree curved path (here $y=x^3$, matching the $x^6$ in the denominator) can still disagree.
• Treating "the limit along one path equals $f(a,b)$" as proof of continuity. Continuity needs the limit to exist (all paths agreeing) and to equal $f(a,b)$; one path is never enough evidence.
• Letting the polar bound secretly depend on $\theta$. Writing something like $|f|\le r/\cos\theta$ and calling it a valid squeeze bound is invalid, since that bound blows up as $\theta\to\pi/2$ instead of shrinking uniformly with $r$.
• Assuming every rational function of $x,y$ is continuous everywhere. It is continuous only where the denominator is nonzero; at zeros of the denominator the limit must be investigated separately, exactly as in the worked examples above.

Ask the tutor: if you want to test whether a specific two-variable limit exists, or practice choosing the right second path or polar bound, ask the AI tutor to work through it with you step by step.