1.2 Limits and Continuity
Definition 1.2.1. Let \(f\) be defined on a set \(D\) containing points arbitrarily close to \((x_0,y_0)\). Then \[\lim _{(x,y) \to (x_0,y_0)}f(x,y) = L\] means: for every \(\varepsilon >0\) there is a \(\delta >0\) such that \[|f(x,y) - L| < \varepsilon \quad \text {whenever}\quad (x,y)\in D\ \text {and}\ 0 < \sqrt {(x-x_0)^2 + (y-y_0)^2} <\delta .\]
In words, \(f(x,y)\) can be made as close to \(L\) as we please by taking \((x,y)\) inside a small enough disc about \((x_0,y_0)\), the centre itself excluded. The condition says nothing about how \((x,y)\) approaches \((x_0,y_0)\): every point of the disc counts. That is what makes limits in the plane more delicate than on the line.
Definition 1.2.2. The function \(f\) is continuous at the point \(P_0(x_0,y_0)\) if all three of the following hold.
- 1.
- \(f\) is defined at \(P_0\); that is, \(f(x_0,y_0)\) exists.
- 2.
- The limit \(\lim _{(x,y)\to (x_0,y_0)} f(x,y)\) exists.
- 3.
- The two agree: \(\lim _{(x,y)\to (x_0,y_0)} f(x,y) = f(x_0,y_0)\).
The function \(f\) is continuous on a set \(S\) if it is continuous at every point of \(S\).
The three conditions are independent, and a single example fails each of them in turn: \(f(x,y) = \frac {xy}{x^2+y^2}\) is undefined at the origin and so fails the first; assigning it any value there still fails the second, since the limit does not exist; and a function whose limit at \(P_0\) exists but differs from its value there fails only the third. This is the same three-part test as for one variable. What is not the same is condition 2, and the difference is the whole difficulty: on the line a point can be approached from two sides, in the plane from infinitely many directions, and along curves as well as straight lines.
Example 1.2.3. Let \[f(x,y) = \begin {cases} 3xy, & (x,y) \neq (1,2),\\ 0, & (x,y) = (1,2). \end {cases}\] Then \(f\) is defined at \((1,2)\), and since \(3xy\) is a polynomial, \(\lim _{(x,y)\to (1,2)}f(x,y) = 3\cdot 1\cdot 2 = 6\). But \(f(1,2) = 0\neq 6\), so condition 3 fails and \(f\) is not continuous at \((1,2)\).
Redefining the value there as \(f(1,2) = 6\) makes the limit and the value agree, and the new function, which is just \(3xy\) everywhere, is continuous. A discontinuity that can be removed in this way, by changing a single value, is called removable.
Example 1.2.4. Let \(f(x,y) = \frac {xy}{x^2 + y^2}\) for \((x,y) \neq (0,0)\). Does \(f\) have a limit as \((x,y) \to (0,0)\)?
Solution. Approach the origin along the line \(y = mx\). For \(x\neq 0\), \[f(x,mx) = \frac {x(mx)}{x^2 + (mx)^2} = \frac {m x^2}{x^2(1 +m^2)} = \frac {m}{1 + m^2},\] so on each line through the origin \(f\) is constant, and the constant depends on the line. Along \(y = x\) it is \(\frac {1}{2}\); along \(y = 2x\) it is \(\frac {2}{5}\). If the limit existed and equalled \(L\), every path would have to give \(L\). Two paths give different values, so the limit does not exist.
The argument is worth isolating, since it is the standard tool for showing that a limit does not exist.
Proposition 1.2.5 (Two-path test). If \(f(x,y)\to L_1\) as \((x,y)\to (x_0,y_0)\) along one curve and \(f(x,y)\to L_2\) along another, with \(L_1\neq L_2\), then \(\lim _{(x,y)\to (x_0,y_0)} f(x,y)\) does not exist.
Proof. If the limit existed and equalled \(L\), then for any \(\varepsilon >0\) all points of \(D\) in some punctured disc about \((x_0,y_0)\) would give values within \(\varepsilon \) of \(L\), including the points of both curves. Hence both \(L_1\) and \(L_2\) would equal \(L\). □
The test can only ever prove that a limit fails to exist. To prove that a limit does exist, a bound valid for all points near \((x_0,y_0)\) is needed, and the usual device is to compare with the distance \(r = \sqrt {(x-x_0)^2 + (y-y_0)^2}\), as in the next example.
Example 1.2.6. Investigate the continuity at the origin of
- 1.
- \(z = \frac {\sin (x+y)}{x + y}\), using \(\lim _{t\to 0} \frac {\sin t}{t}=1\);
- 2.
- \(z = \frac {xy}{x^2 + y^2}\).
Solution.
- 1.
- The function is undefined wherever \(x + y = 0\), which includes the origin, so it is not continuous there as it stands. Its limit does exist: with \(t = x + y\), which tends to \(0\) as \((x,y)\to (0,0)\), \[\lim _{(x,y)\to (0,0)}\frac {\sin (x+y)}{x+y} = \lim _{t\to 0}\frac {\sin t}{t} = 1 .\] So the discontinuity is removable: defining \(z = 1\) on the line \(x + y = 0\) gives a function continuous everywhere.
- 2.
- By Example 1.2.4 the limit at the origin does not exist, so no choice of value at the origin can make this function continuous there.
Extension to Three or More Variables
For a function of three variables, \(\lim _{(x,y,z)\to (a,b,c)}f(x,y,z) =L\) means that \(|f(x,y,z) - L|<\varepsilon \) whenever \(0 < \sqrt {(x-a)^2 + (y-b)^2 + (z-c)^2} < \delta \): the disc is replaced by a ball. Continuity is defined by the same three conditions, and the same rules hold. Sums, products and quotients of continuous functions are continuous where the denominator is non-zero, and so are compositions of continuous functions. In particular every polynomial is continuous everywhere, and every rational function is continuous wherever its denominator is non-zero.
For example, \[f(x,y,z) = \frac {1}{x^2 + y^2 + z^2 - 1}\] is a rational function of three variables, continuous everywhere in \(\mathbb {R}^3\) except where \(x^2 + y^2 + z^2 = 1\): it is discontinuous exactly on the sphere with centre \((0,0,0)\) and radius \(1\).
Solution. Every straight line through the origin gives \(0\), but that proves nothing, so a bound is needed. Since \(x^2\leq x^2+y^2\), \[\left |\frac {3x^2y}{x^2+y^2}\right | = 3|y|\,\frac {x^2}{x^2+y^2}\leq 3|y|\leq 3\sqrt {x^2+y^2} .\] Given \(\varepsilon >0\), take \(\delta = \frac {\varepsilon }{3}\). Then \(0<\sqrt {x^2+y^2}<\delta \) gives \(|f(x,y) - 0|<3\delta = \varepsilon \), so the limit is \(0\). Defining \(f(0,0) = 0\) makes \(f\) continuous at the origin.
Solution. Polar coordinates turn the question into one about a single variable. With \(x = r\cos \theta \) and \(y = r\sin \theta \), \[\frac {x^2-y^2}{\sqrt {x^2+y^2}} = \frac {r^2(\cos ^2\theta - \sin ^2\theta )}{r} = r\cos 2\theta .\] Since \(|r\cos 2\theta |\leq r\) for every \(\theta \), the expression tends to \(0\) as \(r\to 0\), whatever the direction, curved or straight. The limit exists and equals \(0\). The method works because the bound \(r\) does not depend on \(\theta \); had the result been \(\cos 2\theta \) alone, with no factor \(r\), the limit would have failed to exist.
Example 1.2.9. Show that \(\lim _{(x,y)\to (0,0)}\frac {x^2y}{x^4+y^2}\) does not exist, even though the function tends to \(0\) along every straight line through the origin.
Solution. Along the line \(y = mx\) with \(m\neq 0\), \[\frac {x^2(mx)}{x^4 + m^2x^2} = \frac {mx}{x^2+m^2}\to 0 ,\] and along both axes the function is identically \(0\). Along the parabola \(y = x^2\), however, \[\frac {x^2\cdot x^2}{x^4 + x^4} = \frac 12\quad \text {for every } x\neq 0 .\] By Proposition 1.2.5 the limit does not exist. The parabola is the natural path to try because it makes the two terms \(x^4\) and \(y^2\) of the denominator the same size.
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