1.2 Limits and Continuity
The function \(f(x,y)\) is said to have a limit at the point \(P_0(x_0,y_0)\) and we can write \[ \lim \limits _{(x,y) \longrightarrow (x_0,y_0)}f(x,y) = L\] If for each positive \(\varepsilon \), there exists a \(\delta \) that if \[0 < (x-x_0)^2 + (y-y_0)^2 <\delta ^2, \hspace {0.3cm}\text {then}\hspace {0.3cm} \big |f(x,y) - L\big | < \varepsilon \]
i.e the point \(P(x,y)\) lies inside a certain circle of radius \(\delta \), and centre \(P_0(x_0,y_0)\).
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 \(\displaystyle {\lim _{(x,y)\rightarrow (x_0,y_0)} f(x,y)}\) exists.
- 3.
- The two agree: \(\displaystyle {\lim _{(x,y)\rightarrow (x_0,y_0)} f(x,y) = f(x_0,y_0)}\).
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: \(f(x,y) = \dfrac {xy}{x^2+y^2}\) 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.
\(f(x,y) = \begin {cases} 3xy & (x,y) \neq (1,2)\\\\ 0 & (x,y) = (1,2)\\ \end {cases} \)
- 1.
- \(f(x,y)\) is defined at \((1,2)\)
- 2.
- \(\displaystyle {\lim \limits _{(x,y)\rightarrow (1,2)}f(x,y) = 6}\)
- 3.
- \(\displaystyle { f(1,2) = 0\neq \lim \limits _{(x,y)\rightarrow (1,2)}f(x,y)}\)
Condition 3 is violated
But \(f(x,y) = \begin {cases} 3xy & (x,y) \neq (1,2)\\\\ 6 & (x,y) = (1,2)\\ \end {cases} \)
\(\lim \limits _{(x,y)\rightarrow (1,2)}f(x,y) = 6 = f(1,2)\). Then \(f(x,y)\) is continuous.
Let \(\displaystyle {f(x,y) = \frac {xy}{x^2 + y^2}\hspace {0.1cm}, \hspace {0.4cm} (x,y) \neq (0,0)}\). Does the function have a limit as \(P \longrightarrow (0,0)\)?
Solution.
Denote \(f(P)\) for \(f(x,y)\). Let \(P\) approach \((0,0)\) along the line \(y = mx\)
\begin {align*} f(P) & = \frac {x(mx)}{x^2 + (mx)^2}\hspace {1cm} x\neq 0\\\\ & = \frac {m x^2}{x^2(1 +m^2)}\\\\ & = \frac {m}{1 + m^2} \end {align*}
Hence on each line through origin \(f(p)\) is constant.
For \(y = x\), \(m = 1\) and \(f(P) = \dfrac {1}{2}\) as \(P \longrightarrow (0,0)\)
For \(y = 2x\), \(m = 2\) and \(f(P) = \dfrac {2}{5}\) as \(P \longrightarrow (0,0\)
Since \(\dfrac {1}{2} \neq \dfrac {2}{5}\) , no unique \(L\) that will define 1. Hence the function does not have a limit (or does not exist).
Given the functions
- 1.
- \(\displaystyle {z = \frac {\sin (x+y)}{x + y}}\hspace {0.8cm}\) hint: \(\displaystyle {\lim _{x\longrightarrow 0} \frac {\sin x}{x}=1}\)
- 2.
- \(\displaystyle {z = \frac {xy}{x^2 + y^2}}\)
Investigate whether continuous at the origin.
Extension to Three or More Variables
\(\displaystyle {\lim _{(x,y,z)\rightarrow (a,b,c)}f(x,y,z) =L}\)
Means the values of \(f(x,y,z)\) approach \(L\) as the point approaches \((a,b,c)\).
For example, the function \[f(x,y,z) = \frac {1}{x^2 + y^2 + z^2 - 1}\] is a rational function of three variables and it is continuous everywhere in \(\mathbb {R}^3\) except when \(x^2 + y^2 + z^2 = 1\). It is discontinuous on the sphere with centre \((0,0,0)\) and radius 1.
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