Calculus III 13.02 Limits and Continuity

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13.02 Limits and Continuity

  • The definition for a neighborhood in the plane.
  • The limit for a function with two variables.
  • Extend continuity to a function with two variables.
  • Extend continuity to a function with three variables.

Neighborhoods in the Plane

A neighborhood in the plane is a region in the \(xy\)-plane around some given point. This is analogous to extending an interval on the real number line from one dimension to two dimensions. Using the formula for the distance between two points

\((x,y)\) and \((x_{0},y_{0})\)

in the plane is defined as the \(\delta \)-neighborhood about \((x_{0},y_{0})\) to be the disk centered at \((x_{0},y_{0})\) with radius \(\delta > 0\)

$$\left\{ (x,y): \sqrt{(x-x_{0})^{2} + (y-y_{0})^{2}} < \delta \right\} \:\:\:\: \color{red}{ \text{Open disk }} $$

as shown in Figure 13.2.1. When this formula contains the less than inequality sign, \(<\), the disk is called open, and when it contains the less than or equal to inequality sign, \(\leqslant\), the disk is called closed. This corresponds to the using \(<\) and \(\leqslant\) to define open and closed intervals.

An open disk.
Figure 13.2.1
The boundary and interior points for region \(R\).
Figure 13.2.2

A point \((x_{0},y_{0})\) in a plane region \(R\) is an interior point in \(R\) if there exists a \(\delta\)-neighborhood about \((x_{0},y_{0})\) that lies entirely in \(R\), as shown in Figure 13.2.2. If every point in \(R\) is an interior point, then \(R\) is an open region. A point \((x_{0},y_{0})\) is a boundary point in \(R\) if every open disk centered at \((x_{0},y_{0})\) contains points inside \(R\) and points outside \(R\). By definition, a region must contain its interior points, but it need not contain its boundary points. If a region contains all its boundary points, then the region is closed. A region that contains some but not all its boundary points is neither open nor closed.

Limit for a Function with Two Variables

Definition 13.2.1 The Limit for a Function with Two Variables

For any \((x,y)\) in the disk with radius \(\delta\), the value \(f(x,y)\) lies between \(L+\varepsilon \) and \(L-\varepsilon \).
Figure 13.2.3

Let \(f\) be a function with two variables defined, except possibly at \((x_{0},y_{0})\), on an open disk centered at \((x_{0},y_{0})\), and let \(L\) be a real number. Then

$$ \lim_{(x_{0},y_{0}) \rightarrow (x_{0},y_{0})} f(x,y) = L$$

if for each \(\varepsilon > 0 \) there corresponds a \(\delta > 0\) such that

\( |f(x,y)-L| < \varepsilon\) whenever \(0 < \sqrt{(x-x_{0})^{2} + (y-y_{0})^{2}} < \delta \).

Graphically, the definition for a function limit with two variables implies that for any point \((x,y) \ne (x_{0},y_{0})\) in the disk with radius \(\delta\), the value \(f(x,y)\) lies between \(L+\varepsilon \) and \(L-\varepsilon \), as shown in Figure 13.2.3. The limit for a function with one variable easily extends too the limit for a function with two variables, yet there is a critical difference. To determine whether a function with a single variable has a limit, testing the approach from two directions—from the right and from the left, is needed. When the function approaches the same limit from the right and from the left, the limit exists. For a function with two variables, however, the statement

\((x,y) \rightarrow (x_{0},y_{0})\)

means the point \((x,y)\) is allowed to approach \((x_{0},y_{0})\) from any direction. If the value for

$$ \lim_{(x_{0},y_{0}) \rightarrow (x_{0},y_{0})} f(x,y)$$

is not the same for all possible approaches, or paths, to \((x_{0},y_{0})\), then the limit does not exist.

Example 13.2.1 Verifying a Limit by the Definition

Show that

$$ \lim_{(x_{0},y_{0}) \rightarrow (x_{0},y_{0})} x =a$$

Solution Let \( f(x,y) = x \) and \(L=a\). Show that for each \(\varepsilon >0\), there exists a \(\varepsilon \)-neighborhood about \((a,b)\) such that

\( |f(x,y)-L| = |x-a| < \varepsilon \)

whenever \( (x,y) \ne (a,b)\) lies in the neighborhood. Given the inequality

\( 0 < \sqrt{(x-x_{0})^{2} + (y-y_{0})^{2}} < \delta \)
\( |f(x,y)-a | \) \(=|x-a| \)
\(= \sqrt{(x-a)^{2}}\)
\( \leqslant \sqrt{(x-a)^{2} + (y-b)^{2}} \)
\( < \delta \)

Let \( \delta = \varepsilon \) and the limit is verified.

Example 13.2.2 Verifying a Definite Limit for Using Products and Sums

Evaluate

$$ \lim_{(x,y) \rightarrow (1,2)} \frac{5x^{2}y}{x^{2} + y^{2}}.$$

Solution Applying the limits for products and sums produces

$$ \lim_{(x,y) \rightarrow (1,2)} 5x^{2}y = 5(1^{2})(2) = 10$$

and

$$ \lim_{(x,y) \rightarrow (1,2)} x^{2} + y^{2} = (1^{2}+2^{2}) = 5.$$

Applying the quotient for limits, and the denominator is not 0, produces

$$ \lim_{(x,y) \rightarrow (1,2)} \frac{5x^{2}y}{x^{2} + y^{2}} = \frac{10}{5} = 2.$$

Example 13.2.3 Verifying a Definite Limit When the Denominator is Zero

Figure 13.2.4

Evaluate

$$ \lim_{(x,y) \rightarrow (0,0)} \frac{5x^{2}y}{x^{2} + y^{2}}.$$

Solution Limits for products and sums fail here because both the numerator and denominator are zero. From the graph in Figure 13.2.4 zero might be a reasonable limit. Let \(L=0\) and apply the definition. Given that

\( |y| \leqslant \sqrt{x^{2} + y^{2}} \)

and

$$ \frac{x^{2}}{x^{2} + y^{2}} \leqslant 1. $$

In the neighborhood about \((0,0)\),

\( 0 < \sqrt{x^{2} + y^{2}}< \delta \)

this yields that for \( (x,y) \ne (0,0) \),

\( |f(x,y) - 0|\) $$= \left|\frac{5x^{2}y}{x^{2} + y^{2}} \right|$$
$$= 5|y| \left( \frac{x^{2}}{x^{2} + y^{2}} \right)$$
\( \leqslant 5|y| \)
\( < 5 \delta \)

By choosing \( \delta = \varepsilon /5 \), the limit is

$$ \lim_{(x,y) \rightarrow (0,0)} \frac{5x^{2}y}{x^{2} + y^{2}} = 0.$$

Example 13.2.4 A Limit That Does Not Exist

Figure 13.2.5

Figure 13.2.6

For some functions, it is easy to recognize that a limit does not exist. For instance, it is clear the limit

$$ \lim_{(x,y) \rightarrow (0,0)} \frac{1}{x^{2} + y^{2}} $$

does not exist because the values for \(f(x,y)\) increase without bound as \((x,y)\) approaches \((0,0)\) along any path, as shown in Figure 13.2.5. For other functions, recognizing that a limit does not exist is not easy. Sometimes a limit does not exist because the function approaches different values along different paths.

Show that the limit does not exist.

$$ \lim_{(x,y) \rightarrow (0,0)} \left( \frac{x^{2} - y^{2}}{x^{2} + y^{2}} \right)^{2}$$

Solution The domain for the function

$$ f(x,y) = \left( \frac{x^{2} - y^{2}}{x^{2} + y^{2}} \right)^{2}$$

is all the points in the \(xy\)-plane except for the point \((0,0)\). To show that the limit as \((x,y)\) approaches \((0,0)\) does not exist, consider approaching \((0,0)\) along two different paths, as shown in Figure 13.2.5. Along the \(x\)-axis, every point has the form

\((x,0)\)

and the limit along this approach is

$$ \lim_{(x,0) \rightarrow (0,0)} \left( \frac{x^{2} - 0^{2}}{x^{2} + 0^{2}} \right)^{2} = \lim_{(x,0) \rightarrow (0,0)} 1^{2} = 1. \:\:\:\: \color{red}{ \text{Limit along the } x \text{-axis}} $$

However, when \((x,y)\) approaches \((0,0)\) along the line \(y=x\), the limit is

$$ \lim_{(x,x) \rightarrow (0,0)} \left( \frac{x^{2} - x^{2}}{x^{2} + x^{2}} \right)^{2} = \lim_{(x,x) \rightarrow (0,0)} \left( \frac{0}{2x^{2}} \right)^{2} = 0. \:\:\:\: \color{red}{ \text{Limit along the line } y=x } $$

This means that in any open disk centered at \((0,0)\), there are points \((x,y)\) at which \(f\) takes on the value 1, and other points at which \(f\) takes on the value 0. For instance,

\( f(x,y) = 1\)

at \((1,0)\), \((0.1,0)\), \((0.01,0)\), and \((0.001,0)\), and

\( f(x,y) = 0\)

at \((1,1)\), \((0.1,0.1)\), \((0.01,0.01)\), and \((0.001,0.001)\). Therefore, \(f\) does not have a limit as \((x,y)\) approaches \((0,0)\).

The limit does not exist because two approaches produced different limits or the same limit. The limit must be the same along all possible approaches. One difference on one approach negates any limit.

Continuity for a Function with Two Variables

Notice in Example 13.2.2 that the limit can be evaluated by direct substitution. That is, the limit is \(f(1,2)=2\). In such cases the function \(f\) is said to be continuous at the point \((1,2)\).

Definition 13.2.2 Continuity for a Function with Two Variables

A function \(f\) with two variables is continuous at a point \((x_{0},y_{0})\) in an open region \(R\) if \(f(x_{0},y_{0})\) is equal to the limit for \(f(x,y)\) as \((x,y)\) approaches \((x_{0},y_{0})\). As an equation.

$$ \lim_{(x,y) \rightarrow (x_{0},y_{0})} f(x,y) = f (x_{0},y_{0}).$$

The function \(f\) is continuous in the open region \(R\) if it is continuous at every point in \(R\).

In Example 13.2.3 the function

$$ f(x,y) = \frac{5x^{2}y}{x^{2}+y^{2}}$$

is not continuous at \((0,0)\). Because the limit exists at this point, the discontinuity can be removed by defining \(f\) at \((0,0)\) as being equal to its limit there. Such a discontinuity is called removable. In Example 13.2.4 the function

$$ f(x,y) = \left( \frac{x^{2}-y^{2}}{x^{2}+y^{2}} \right)^{2}$$

was also not continuous at \((0,0)\), but this discontinuity is nonremovable.

Theorem 13.2.1 Continuous Functions with Two Variables

If \(k\) is a real number and \(f\) and \(g\) are continuous at \((x_{0},y_{0})\), then the following functions are also continuous at \((x_{0},y_{0})\).

1. Scalar multiple: \(kf \:\:\:\: \) 2. Sum or difference: \(f \pm g \)
3. Product: \(fg\) 4. Quotient: \(f/g, \: g(x_{0},y_{0}) \ne 0\)

Theorem 13.2.1 establishes the continuity for polynomial and rational functions at every point in their domains. Furthermore, the continuity for other functions types can be extended naturally from one to two variables. For instance, the functions whose graphs are shown in Figures 13.2.7 and 13.2.8 are continuous at every point in the plane.

Figure 13.2.7
Figure 13.2.8

Theorem 13.2.2 Continuity for a Composite Function

If \(h\) is continuous at \((x_{0},y_{0})\) and \(g\) is continuous at \(h(x_{0},y_{0})\), then the composite function given by \((g \circ h)(x,y) =g(h((x,y))\) is continuous at \((x_{0},y_{0})\). That is,

\(\lim_{(x,y) \rightarrow (x_{0},y_{0})} g(h((x,y)) = g(h(x_{0},y_{0})\).

Note that \(h\) is a function with two variables and \(g\) is a function with one variable.

Example 13.2.5 Testing for Continuity

Discuss the continuity for each function.

$$\textbf{a. } f(x,y) = \frac{x-2y }{x^{2}+y^{2}}\:\:\:\:\textbf{b. } g(x,y) = \frac{2 }{y-x^{2}} $$

Solution a. Because a rational function is continuous at every point in its domain, you can conclude that is continuous at each point in the \(xy\)-plane except at \((0,0)\), as shown in Figure 13.2.9.
b. The function

$$g(x,y) = \frac{2 }{y-x^{2}} $$

is continuous except at the points at which the denominator is zero, which is given by the equation

\(y-x^{2} = 0\).

The conclusion is that \(g\) is continuous at all points except those lying on the parabola \(y=x^{2}\). Inside this parabola \(y>x^{2}\), and the surface for \(g\) lies above the \(xy\)-plane, as shown in Figure 13.2.10. Outside the parabola, \(y<x^{2}\), and the surface lies below the \(xy\)-plane.

The function \(f\) is not continuous at \((0,0)\).
Figure 13.2.9
The function \(g\) is not continuous on the parabola \(y=x^{2}\).
Figure 13.2.10

Continuity for a Function with Three Variables

Open sphere in three-dimensions
Figure 13.2.11

Definitions 13.2.1 and 13.2.2 can be extended to functions with three variables by considering point \((x,y,z)\) within an open sphere

\( (x-x_{0})^{2} + (y-y_{0})^{2} + (z-z_{0})^{2} < \delta^{2}. \:\:\:\: \) Open sphere

The radius for this sphere is \(\delta\), and it is centered at \((x_{0},y_{0},z_{0})\), as shown in Figure 13.2.11. A point \((x_{0},y_{0},z_{0})\) in a region \(R\) in three-dimensions is an interior point for \(R\) if there exists a \(\delta\)-sphere about \((x_{0},y_{0},z_{0})\) that lies entirely in \(R\). IF every point in \(R\) is an interior point, then \(R\) is called open.

Definition 13.2.3 Continuity for a Function with Three Variables

A function \(f\) with three variables is continuous at a point \((x_{0},y_{0},z_{0})\) in an open region \(R\) if \(f(x_{0},y_{0},z_{0})\) is defined and is equal to the limit for \(f(x,y,z)\) as \((x,y,z)\) approaches \((x_{0},y_{0},z_{0})\). As an equation

$$ \lim_{(x,y,z) \rightarrow (x_{0},y_{0},z_{0})} f(x,y,z) = f (x_{0},y_{0},z_{0}).$$

The function \(f\) is continuous in the open region \(R\) if it is continuous at every point \(R\).

Example 13.2.6 Testing Continuity for a Function with Three Variables

Discuss the continuity for

$$ f(x,y,z) = \frac{1}{ x^{2}+y^{2}-z }. $$

Solution The function \(f\) is continuous except at the points where the denominator is zero, which are given by the equation

\( x^{2}+y^{2}-z = 0 \).

Therefore, \(f\) is continuous at each point in three-dimensions except at the points on the paraboloid

\(z=x^{2}+y^{2}\).
Square X.jpg

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Parent Article: Calculus III 13 Functions with Several Variables