## 2010. Limits and Continuity LIMITS AND CONTINUITY

10/13/2010 LIMITS AND CONTINUITY • Let’s compare the behavior of the functions 15.2 sin( x 2 + y 2 ) x2 − y 2 and g ( x , y ) = x2 + y2 x2 + y2 as ...
Author: Erica Clarke
10/13/2010

LIMITS AND CONTINUITY • Let’s compare the behavior of the functions

15.2

sin( x 2 + y 2 ) x2 − y 2 and g ( x , y ) = x2 + y2 x2 + y2 as x and y both approach 0 (and thus the point (x, y) approaches the origin).

f ( x, y ) =

Limits and Continuity In this section, we will learn about: Limits and continuity of various types of functions. Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMITS AND CONTINUITY

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMITS AND CONTINUITY Table 1 •This table shows values of f(x, y).

• The following tables show values of f(x, y) and g(x, y), correct to three decimal places, for points (x, y) near the origin.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMITS AND CONTINUITY Table 2 •This table shows values of g(x, y).

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMITS AND CONTINUITY • Notice that neither function is defined at the origin. – It appears that, as (x, y) approaches (0, 0), the values of f(x, y) are approaching 1, whereas the values of g(x, y) aren’t approaching any number.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

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LIMITS AND CONTINUITY • It turns out that these guesses based on numerical evidence are correct. • Thus, we write:

LIMITS AND CONTINUITY • In general, we use the notation

lim

( x , y ) → ( a ,b )

f ( x, y ) = L

to indicate that: –

sin( x 2 + y 2 ) =1 ( x , y ) →(0,0) x2 + y2 – does not exist. x2 − y 2 lim ( x , y ) → (0,0) x 2 + y 2 lim

– The values of f(x, y) approach the number L as the point (x, y) approaches the point (a, b) along any path that stays within the domain of f.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMITS AND CONTINUITY • In other words, we can make the values of f(x, y) as close to L as we like by taking the point (x, y) sufficiently close to the point (a, b), but not equal to (a, b).

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION Definition 1

• Let f be a function of two variables whose domain D includes points arbitrarily close to (a, b). • Then, we say that the limit of f(x, y) as (x, y) approaches (a, b) is L.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

SINGLE VARIABLE FUNCTIONS • For functions of a single variable, when we let x approach a, there are only two possible directions of approach, from the left or from the right.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

DOUBLE VARIABLE FUNCTIONS • For functions of two variables, the situation is not as simple.

– We recall from Chapter 2 that, if then lim f ( x) does not exist. lim− f ( x ) ≠ lim+ f ( x), x →a

x→a

x →a

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

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DOUBLE VARIABLE FUNCTIONS • This is because we can let (x, y) approach (a, b) from an infinite number of directions in any manner whatsoever as long as (x, y) stays within the domain of f.

LIMIT OF A FUNCTION • Definition 1 refers only to the distance between (x, y) and (a, b).

– It does not refer to the direction of approach.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION • Therefore, if the limit exists, then f(x, y) must approach the same limit no matter how (x, y) approaches (a, b).

LIMIT OF A FUNCTION • Thus, if we can find two different paths of approach along which the function f(x, y) has different limits, then it follows that does not exist.

lim

( x , y ) → ( a ,b )

f ( x, y )

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION • If f(x, y) → L1 as (x, y) → (a, b) along a path C1 and f(x, y) → L2 as (x, y) → (a, b) along a path C2, where L1 ≠ L2, then

lim

( x , y ) → ( a ,b )

LIMIT OF A FUNCTION Example 1

• Show that

x2 − y2 ( x , y ) → (0,0) x 2 + y 2 lim

does not exist.

f ( x, y )

– Let f(x, y) = (x2 – y2)/(x2 + y2).

does not exist. Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

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LIMIT OF A FUNCTION Example 1

LIMIT OF A FUNCTION Example 1

• First, let’s approach (0, 0) along the x-axis.

• We now approach along the y-axis by putting x = 0.

– Then, y = 0 gives f(x, 0) = x2/x2 = 1 for all x ≠ 0.

– Then, f(0, y) = –y2/y2 = –1 for all y ≠ 0.

– So, f(x, y) → 1 as (x, y) → (0, 0) along the x-axis.

– So, f(x, y) → –1 as (x, y) → (0, 0) along the y-axis.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION • Since f has two different limits along two different lines, the given limit does not exist. – This confirms the conjecture we made on the basis of numerical evidence at the beginning of the section.

LIMIT OF A FUNCTION Example 2

• If

f ( x, y ) =

xy x + y2

lim

f ( x, y )

Example 1

does

( x , y ) →(0,0)

2

exist? Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION Example 2

LIMIT OF A FUNCTION Example 2

• If y = 0, then f(x, 0) = 0/x2 = 0. – Therefore,

• If x = 0, then f(0, y) = 0/y2 = 0. – So,

f(x, y) → 0 as (x, y) → (0, 0) along the x-axis. Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

f(x, y) → 0 as (x, y) → (0, 0) along the y-axis. Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

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LIMIT OF A FUNCTION

LIMIT OF A FUNCTION Example 2

Example 2

• Although we have obtained identical limits along the axes, that does not show that the given limit is 0.

• Let’s now approach (0, 0) along another line, say y = x. – For all x ≠ 0,

f ( x, x ) = – Therefore,

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION Example 2

• Since we have obtained different limits along different paths, the given limit does not exist.

f ( x, y ) → 12 as ( x, y ) → (0, 0) along y = x

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION •This figure sheds some light on Example 2. – The ridge that occurs above the line y = x corresponds to the fact that f(x, y) = ½ for all points (x, y) on that line except the origin.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION

x2 1 = x + x2 2 2

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION

Example 3

• If

does

Example 3

xy 2 f ( x, y ) = 2 x + y4

lim

( x , y ) → (0,0)

• With the solution of Example 2 in mind, let’s try to save time by letting (x, y) → (0, 0) along any nonvertical line through the origin.

f ( x, y )

exist? Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

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LIMIT OF A FUNCTION

LIMIT OF A FUNCTION

Example 3

Example 3

• Then, y = mx, where m is the slope, and f ( x, y ) = f ( x, mx)

=

• Therefore, f(x, y) → 0 as (x, y) → (0, 0) along y = mx

x(mx) 2 x 2 + (mx) 4

– Thus, f has the same limiting value along every nonvertical line through the origin.

m2 x3 x 2 + m4 x 4 m2 x = 1 + m4 x2 =

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION

Example 3

• However, that does not show that the given limit is 0.

Example 3

– This is because, if we now let (x, y) → (0, 0) along the parabola x = y2 we have: y2 ⋅ y2 y4 2

f ( x, y ) = f ( y , y ) =

2 2

(y ) + y

=

4

2y

4

=

1 2

• Since different paths lead to different limiting values, the given limit does not exist.

– So, f(x, y) → ½ as (x, y) → (0, 0) along x = y2 Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION

• Now, let’s look at limits that do exist.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION • Just as for functions of one variable, the calculation of limits for functions of two variables can be greatly simplified by the use of properties of limits.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

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LIMIT OF A FUNCTION • The Limit Laws listed in Section 2.3 can be extended to functions of two variables.

LIMIT OF A FUNCTION Equations 2

• In particular, the following equations are true.

• For instance,

lim

x=a

lim

y=b

lim

c=c

( x , y ) → ( a ,b )

– The limit of a sum is the sum of the limits.

( x , y ) → ( a ,b )

– The limit of a product is the product of the limits. Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

LIMIT OF A FUNCTION Equations 2

( x , y ) → ( a ,b )

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

CONTINUITY OF SINGLE VARIABLE FUNCTIONS • Recall that evaluating limits of continuous functions of a single variable is easy.

• The Squeeze Theorem also holds.

– It can be accomplished by direct substitution. – This is because the defining property of a continuous function is

lim f ( x ) = f (a ) x→a

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

CONTINUITY OF DOUBLE VARIABLE FUNCTIONS • Continuous functions of two variables are also defined by the direct substitution property.

CONTINUITY Definition 4

• A function f of two variables is called continuous at (a, b) if

lim

( x , y ) →( a ,b )

f ( x, y ) = f ( a, b )

• We say f is continuous on D if f is continuous at every point (a, b) in D.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

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CONTINUITY

CONTINUITY

• The intuitive meaning of continuity is that, if the point (x, y) changes by a small amount, then the value of f(x, y) changes by a small amount. – This means that a surface that is the graph of a continuous function has no hole or break.

• Using the properties of limits, you can see that sums, differences, products, quotients of continuous functions are continuous on their domains. – Let’s use this fact to give examples of continuous functions.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

POLYNOMIAL

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

RATIONAL FUNCTION

• A polynomial function of two variables (polynomial, for short) is a sum of terms of the form cxmyn, where:

• A rational function is a ratio of polynomials.

– c is a constant. – m and n are nonnegative integers.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

RATIONAL FUNCTION VS. POLYNOMIAL f ( x, y ) = x 4 + 5 x 3 y 2 + 6 xy 4 − 7 y + 6

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

CONTINUITY • The limits in Equations 2 show that the functions

• is a polynomial.

g ( x, y ) =

f(x, y) = x, g(x, y) = y, h(x, y) = c

2 xy + 1 x2 + y2

are continuous.

• is a rational function.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

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CONTINUOUS POLYNOMIALS • Any polynomial can be built up out of the simple functions f, g, and h by multiplication and addition.

CONTINUOUS RATIONAL FUNCTIONS • Likewise, any rational function is continuous on its domain because it is a quotient of continuous functions.

– It follows that all polynomials are continuous on R2.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

CONTINUITY

CONTINUITY

Example 5 Example 5

• Evaluate

lim ( x 2 y 3 − x 3 y 2 + 3 x + 2 y )

– Hence, we can find the limit by direct substitution:

( x , y ) →(1,2)

lim ( x 2 y 3 − x3 y 2 + 3x + 2 y )

2 3 3 2 – f ( x, y ) = x y − x y + 3 x + 2 y is a polynomial.

( x , y )→ (1,2)

= 12 ⋅ 23 − 13 ⋅ 22 + 3 ⋅1 + 2 ⋅ 2

– Thus, it is continuous everywhere.

= 11 Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

CONTINUITY

CONTINUITY Example 6

Example 6

• The function f is discontinuous at (0, 0) because it is not defined there.

• Where is the function

f ( x, y ) = continuous?

x2 − y 2 x2 + y 2

• Since f is a rational function, it is continuous on its domain, which is the set D = {(x, y) | (x, y) ≠ (0, 0)}

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

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CONTINUITY

CONTINUITY Example 7

• Let

 x2 − y2 if ( x, y ) ≠ (0, 0)  2 g ( x, y ) =  x + y 2 0 if ( x, y ) = (0, 0) 

•This figure shows the graph of the continuous function in Example 8.

– Here, g is defined at (0, 0). – However, it is still discontinuous there because

lim

( x , y ) → (0,0)

g ( x, y )

does not exist (see Example 1). Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

COMPOSITE FUNCTIONS • Just as for functions of one variable, composition is another way of combining two continuous functions to get a third.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

COMPOSITE FUNCTIONS • In fact, it can be shown that, if f is a continuous function of two variables and g is a continuous function of a single variable defined on the range of f, then – The composite function h = g ◦ f defined by h(x, y) = g(f(x, y)) is also a continuous function.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

COMPOSITE FUNCTIONS Example 9

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

COMPOSITE FUNCTIONS Example 9

• Where is the function h(x, y) = arctan(y/x) continuous?

–So, the composite function g(f(x, y)) = arctan(y/ x) = h(x, y)

– The function f(x, y) = y/x is a rational function and therefore continuous except on the line x = 0.

is continuous except where x = 0.

– The function g(t) = arctan t is continuous everywhere. Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

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COMPOSITE FUNCTIONS Example 9

•The figure shows the break in the graph of h above the y-axis.

15.3 Partial Derivatives In this section, we will learn about: Multivariable Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

f x (a, b ) = lim h →0

f (a + h, b ) − f (a, b ) h

Partial derivative of f with respect to x at (a, b )

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

f y (a, b ) = lim h →0

Partial derivative of f with respect to y at (a, b )

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

f x ( x, y ) = lim h →0

f ( x + h, y ) − f (x, y ) h

f (a, b + h ) − f (a, b ) h

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

f y ( x, y ) = lim h →0

f ( x, y + h ) − f ( x, y ) h

Partial derivative of f with respect to x as a function itself

Partial derivative of f with respect to y as a function itself

Regard y as a constant and differentiate f ( x, y ) with respect to x

Regard x as a constant and differentiate f (x, y ) with respect to y

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

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Notation:

Clairaut’s Theorem

f x ( x, y ) = f x =

∂f ∂ ∂z = f ( x, y ) = = Dx f ∂x ∂x ∂x

Mixed partials are equal.

∂2 f ∂y∂x The derivative with respect to x first, then the derivative with respect to y of that. f xy ( x, y ) = f xy =

3 Classical Partial Differential Equations ( PDEs )

2

∂ f ∂x 2 The derivative with respect to x first, f xx ( x, y ) = f xx =

then the derivative with respect to x of that.

f xy ( x, y ) = f yx (x, y )

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

f ( x, y ) = 3 x 2 y + y 3 − 3 x 2 − 3 y 2 + 2

ut = ku xx

a 2u xx = utt

u xx + u yy = 0

Heat Equation

Wave Equation

Laplace’s Equation

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

f x = 6 xy − 6 x

f y = 3x 2 + 3 y 2 − 6 y

f x (1,1) is the slope

f xx = 6 y − 6

f yy = 6 y − 6

of the red line.

f xy = 6 x

f yx = 6 x

x g ( x, y ) = x 2 y 3 + ln    y 11 1 3 g x = 2 xy + x   = 2xy 3 + x y  y g xx = 2 y 3 −

1 x2 g xy = g yx = 6 xy 2

1  −x  1 g y = 3 x 2 y 2 + x  2  = 3x 2 y 2 − y y  y  1 2 g yy = 6 x y + 2 y Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives

f y (1,1) is the slope of the red line.

Math 114 – Rimmer 15.2 and 15.3 Limits, Continuity, and Partial Derivatives 71

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