Limits and Continuity in Two Variables

By the end you should be able to: Explain why a limit in two variables must be independent of the path of approach, use the two-path test to show a limit does not exist, and state continuity for a function of two variables.

In one variable, $x$ could approach $a$ from two directions. Checking both was enough. In two variables there are infinitely many directions — and curved paths besides.

That single change is what makes multivariable limits harder, and it is the whole content of this topic.

The definition

\lim_{(x,y)\to(a,b)} f(x,y) = L

means $f(x,y)$ can be made as close to $L$ as we like by taking $(x,y)$ close enough to $(a,b)$ — along any path whatsoever.

Formally: for every $\varepsilon>0$ there is $\delta>0$ with

0 < \sqrt{(x-a)^2+(y-b)^2} < \delta \implies |f(x,y)-L| < \varepsilon

The distance is now Euclidean rather than $|x-a|$ , and "close" means inside a small disc rather than a small interval. Everything else is as before.

The two-path test

Proving a limit exists requires it to hold along every path — hard. Proving one does not exist is much easier:

Find two paths giving different values. Done.

If the answer depends on how you approach, there is no single limit.

The standard example

f(x,y) = \frac{xy}{x^2+y^2} \quad \text{as } (x,y) \to (0,0)

Approach along any straight line through the origin, $y = mx$ :

f(x,mx) = \frac{x(mx)}{x^2+(mx)^2} = \frac{mx^2}{x^2(1+m^2)} = \frac{m}{1+m^2}

The $x$ has cancelled entirely. The value depends only on the slope of the line you came in on:

Path	$m$	Limit along it
along the $x$ -axis	0	$0$
along $y = x$	1	$\tfrac12$
along $y = -x$	$-1$	$-\tfrac12$
along $y = 2x$	2	$\tfrac25$

Different answers, so $\lim_{(x,y)\to(0,0)} f(x,y)$ does not exist.

Geometrically the surface is a ridge that spirals: near the origin it takes every value between $-\tfrac12$ and $\tfrac12$ , no matter how close you look.

The trap

Approaching along the $x$ -axis gives 0. Along the $y$ -axis, also 0. A student who checks only those two concludes the limit is 0 — and is wrong.

Two paths agreeing proves nothing. Only disagreement is conclusive. To establish existence you need an argument covering all paths at once, such as the squeeze theorem or a bound in polar coordinates.

Continuity

Identical in form to the one-variable case:

f \text{ is continuous at } (a,b) \iff \lim_{(x,y)\to(a,b)} f(x,y) = f(a,b)

with the same three requirements — $f(a,b)$ defined, the limit exists, the two agree.

A discontinuity is now a hole, tear or fold in a surface rather than a gap in a curve. Polynomials in $x$ and $y$ are continuous everywhere; rational functions are continuous wherever the denominator is non-zero.