Limits and Continuity

In one variable, $x$ can approach $a$ from only two sides, and the precise definition of a limit asks for closeness on both. In the plane, a point can approach $(a, b)$ along infinitely many paths: straight lines from every direction, parabolas, spirals. A limit exists only if the function settles on one value no matter how you approach.

Key idea

$\displaystyle\lim_{(x, y) \to (a, b)} f(x, y) = L$ means: for every $\varepsilon > 0$ there is a $\delta > 0$ such that $|f(x, y) - L| < \varepsilon$ whenever ${}0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta$.
To show a limit does not exist: find two paths that give different values. To show it exists: squeeze it, often with polar coordinates $x = r\cos\theta$, $y = r\sin\theta$, where $(x, y) \to (0, 0)$ means $r \to 0$ for every $\theta$.

Why it works

The definition talks about every point in a small disk around $(a, b)$, not about any particular path. If the limit is $L$, then any path into $(a, b)$ eventually stays inside the disk, so every path gives $L$. Turned around: two paths with different values make a limit impossible. But checking many paths can never prove a limit exists, because there are always paths you have not tried. A proof has to control all points at once, and a bound like $|f(x, y) - L| \le C r$ does exactly that.

$z = \frac{2x^2y}{x^4 + y^2}$. Along every straight line into the origin the height goes to ${}0$, but a ridge of height ${}1$ follows the parabola $y = x^2$ right into the origin, and a valley of depth $-1$ follows $y = -x^2$.

Worked examples

Worked example: Lines disagree

Show that $\displaystyle\lim_{(x, y) \to (0, 0)} \frac{xy}{x^2 + y^2}$ does not exist.

Along $y = mx$ (with $x \ne 0$): $\frac{x \cdot mx}{x^2 + m^2x^2} = \frac{m}{1 + m^2}$.
$m = 0$ (the $x$-axis) gives ${}0$; $m = 1$ gives $\frac{1}{2}$.
Two paths, two values: no limit.

Answer: does not exist

Worked example: Lines agree, but a parabola does not

Does $\displaystyle\lim_{(x, y) \to (0, 0)} \frac{x^2y}{x^4 + y^2}$ exist?

Along $y = mx$: $\frac{mx^3}{x^4 + m^2x^2} = \frac{mx}{x^2 + m^2} \to 0$ for every $m$ (and the $y$-axis gives ${}0$ too).
Along $y = x^2$: $\frac{x^4}{x^4 + x^4} = \frac{1}{2}$ for every $x \ne 0$.
Different values on two paths, so the limit does not exist.

Answer: does not exist

Worked example: A squeeze in polar coordinates

Find $\displaystyle\lim_{(x, y) \to (0, 0)} \frac{x^2y}{x^2 + y^2}$.

With $x = r\cos\theta$, $y = r\sin\theta$: $\frac{r^3\cos^2\theta\sin\theta}{r^2} = r\cos^2\theta\sin\theta$.
$|\cos^2\theta\sin\theta| \le 1$, so $|f(x, y)| \le r$.
As $(x, y) \to (0, 0)$, $r \to 0$, so $f \to 0$ along every path at once.

Answer: ${}0$

Polynomials are continuous everywhere, and so are rational functions wherever the denominator is not zero; sums, products and compositions of continuous functions are continuous. For those, the limit is just the value: substitute.

Common mistake

Checking only straight lines. In the second example every line gives ${}0$, yet the limit does not exist: the parabola $y = x^2$ gives $\frac{1}{2}$. Agreement along lines is evidence, not proof. Either find a path that disagrees (try $y = x^2$ or $x = y^2$ when the powers in the denominator differ) or prove the limit with a bound that holds for all points near the origin.

Limits and Continuity

Worked examples

Practice

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