1.2 Algebraic Limits And Continuity

Math 135 Class Notes

1.2

Business Calculus

Spring 2009

Algebraic Limits and Continuity

In the last section, we showed how we can evaluate limits numerically and graphically. However, these methods can be time consuming and are often not practical for finding the exact values of limits. In this section, we’ll see how to evaluate limits algebraically using the following principles. LIMIT PRINCIPLES If lim f (x) = L and lim g(x) = M , then we have the folowing: x→a x→a 1.1 lim c = c. x→a

(The limit of a constant is the constant.) 1.2 For any positive integer n,

in £ §n h lim f (x) = lim f (x) = Ln ,

x→a

and

lim

x→a

x→a

q p √ n n f (x) = n lim f (x) = L, x→a

assuming that L ≥ 0 when n is even. (The limit of a power is the power of the limit, and the limit of a root is the root of the limit.) £ § 1.3 lim f (x) ± g(x) = lim f (x) ± lim g(x) = L ± M . x→a

x→a

x→a

(The limit of a sum or difference is the sum or the difference of the limits.) i h i £ § h 1.4 lim f (x) · g(x) = lim f (x) · lim g(x) = L · M . x→a

x→a

x→a

(The limit of a product is the product of the limits.) lim g(x) g(x) M = x→a = provided that L 6= 0. x→a f (x) lim g(x) L

1.5 lim

x→a

(The limit of a quotient is the quotient of the limits.) £ § ˙ 1.6 lim cf (x) = c · lim f (x) = cL. x→a

x→a

(The limit of a constant times a function is the constant times the limit.)

It is easy to see why these properties are true. For instance, if f (x) is close to L and g(x) is close to M , it is reasonable to conclude that f (x) + g(x) is close to L + M . This gives us an intuitive basis for believing that the limit of a sum is the sum of the limits. EXAMPLE

Use the Limit Principles to find lim (x2 − 3x + 7). x→4

5

6

Chapter 1

Differentiation

Note that in the preceding example, if we let f (x) = x2 − 3x + 7, then lim f (x) = f (4). That is, x→4

the limit of f (x) as x approaches 4 equals the value of the function at x = 4. EXAMPLE

x3 + 3x2 − 1 . x→−2 x+4

Use the Limit Principles to find lim

x3 + 3x2 − 1 , then lim g(x) = g(−2). Again, the limit of x→−2 x+4 the function as x approaches −2 equals the value of the function at x = −2. The result of these two examples can be shown to hold in general, as stated in the following theorem. In this example, also, if we let g(x) =

THEOREM ON LIMITS OF POLYNOMIALS AND RATIONAL FUNCTIONS For any polynomial or rational function F (x), with a in the domain of F , lim F (x) = F (a).

x→a

The limit of a polynomial or rational function as x approaches a equals the value of the function at x = a. p EXAMPLE Find lim x2 − 3x + 2. x→0

1.2

Algebraic Limits and Continuity

x2 − 9 . x→−3 x + 3

EXAMPLE

7

Find lim

x2 − 9 in this limit is a rational function, we cannot apply the x+3 theorem to evaluate the limit as x approaches −3 since −3 is not in the domain of f . a) Numerically by completing the table of values. Note that even though the function

x approaches −3 from left −→ | ←− x approaches −3 from right x f (x)

−4

−3.1

−3.01 −3.001

−3

b) Graphically y 6 5 4 3 2 1 –6 –5 –4 –3 –2 –1

1 –1 –2 –3 –4 –5 –6

c) Algebraically

2

3

4

5

6

x

−2.999 −2.99

−2.9

−2

8

Chapter 1

EXAMPLE

Differentiation

Find lim (3x2 + 3xh + h2 ) h→0

CONTINUITY We saw above that for certain functions, the limit of the function as x approaches a can be found simply by calculating the value of the function at a. Functions with this property are called continuous at a. DEFINITION OF CONTINUOUS FUNCTIONS A function f is continuous at x = a provided that: a) f (a) exists, b) lim f (x) exists, and x→a

c) lim f (x) = f (a). x→a

(The limit is the same as the function value.)

A function f is continuous on an interval I provided that it is continuous at each point in I. EXAMPLE

Determine whether the function given by f (x) = 2x + 3 is continuous at x = 4.

1.2

EXAMPLE

Is the function g defined by g(x) =

continuous at x = −2? Why or why not?

Ω

1 2x

+ 3 for x < −2 x−1 for x ≥ −2

y 6 5 4 3 2 1 –6 –5 –4 –3 –2 –1

1

2

3

4

5

6

x

–1 –2 –3 –4 –5 –6

EXAMPLE

Is the function F defined by

 2 x −9 F (x) = x+3  2 continuous at x = −3? Why or why not? y 6 5 4 3 2 1 –6 –5 –4 –3 –2 –1

1 –1 –2 –3 –4 –5 –6

2

3

4

5

6

x

for x 6= −3 for x = −3

Algebraic Limits and Continuity

9

10

Chapter 1

EXAMPLE

Differentiation

The graph of the function f (x) =

Why or why not?

1 + 3 is shown below. Is f continuous at x = 2? x−2

y 8 7 6 5 4 3 2 1 –4

–3

–2

–1

1

2

3

4

5

6

7

8

x

–1 –2 –3 –4

From the preceding three examples, we see that if the graph of a function has a “jump” or a “hole” or goes off to infinity at a point x = a, then the function will not be continuous at a and hence not continous on any interval containing a. In general, in order for a function to be continuous on an interval, then its graph can be traced without lifting a pencil from the paper. EXAMPLE why?

The figure shows the graph of a function f . At which numbers is f discontinuous and