CONTINUITY, END BEHAVIOR,

AND LIMITS

CONTINUITY A continuous function has no breaks,

holes, or gaps

You can trace a continuous function without lifting your pencil

TYPES OF DISCONTINUITY Infinite (non-removable)

The function value increases or decreases indefinitely as x gets close to the point of discontinuity (there is a vertical asymptote at the point of discontinuity)

Jump (non-removable) The function values differ when approaching the point

of discontinuity from the left and the right (often a piecewise-defined function)

Removable The function approaches the same value from the left

and the right. However, the function is either undefined at the point or not equal to the value the graph is approaching. (there is a hole)

LIMITS

The concept of a limit is important in calculus. It is a value that a function is approaching. The function may or may not ever reach this value.

For a limit to exist at a point c the function must approach the same value when c is approached from the left as when c is approached from the right.

lim f(x) = L See examples on board.

x→c

CONTINUITY TESTf(x) is continuous at a point c if:

1. f(c) exists (is defined)2. the limit of f(x) as x approaches c

exists(f(x) approaches the same value from each side of c)

3. These two values are equal

f(c) = lim f(x) x c

CONFIRMING CONTINUTIYExample 1

Show that f(x) = 2x2 – 3x – 1 is continuous at x = 2.

1. f(2) = 1

The function is defined at x = 2.

2. f(1.999) = .995

f(2.001) = 1.005

The limit appears to exist at x = 2.

3. f(2) = lim f(x)

x2

The limit equals the function value.

Graph to confirm. Note that holes and asymptotes may be difficult to identify on a graphing calculator!

SHOWING DISCONTINUITYDetermine whether the function is continuous for the given value of x. If discontinuous, determine the type of discontinuity.

example 2a) Is f(x) = (x – 2)/(x2- 4) continuous at x= 2?1. f(2) = 0/0 is undefined. So not continuous.2. f(1.999) = .25006

f(2.001) = .24994The lim f(x) appears to be .25. x2

3. Because the limit exists this must be a removable discontinuity.

Example 2b) Is f(x) = (x – 2)/(x2- 4) continuous at x= -2?

1. f(-2) = -4/0 is undefined. So not continuous.

2. f(-1.999) = 1000f(-2.001) = -1000

The function f(x) appears to get very large when x approaches -2 from the right. It appears to get very small when it approaches -2 from the left.

3. Because the function appears to increase/ decrease indefinitely this must be an infinite discontinuity.

Graph this!

Example 3

f(x) = {5x + 4 if x < 2

{2 – x if x ≥ 2

Is this function continuous at x = 2?

1. f(2) = 0, so it exists

2. f(1.999) = 13.995

f(2.001) = -.001

So the limit as x approaches 2 does not exist. So not continuous.

3. This must be a jump discontinuity since the function approaches different value from the left and the right.

Graph this.

INTERMEDIATE VALUE THEOREM If f(x) is a continuous function and a < b,

and there is a value n such that n is between f(a) and f(b), then there is a number c between and b such that f(c)=n.

Corollary: The Location PrincipleIf f(x) is a continuous function and f(a) and f(b) have opposite signs, then there must be a zero between a and b. (Although there could be more than one.)

Can there be a zero if the signs are the same?

APPROXIMATING ZEROS Determine between which consecutive

integers the zeros of the function are located on the given interval.

f(x) = 10x2 – 57x + 63 [0,6]

Answer: between 1 and 2 and 4 and 5

Graph to verify.

END BEHAVIOR Describe the end behavior of each

function. First use the graph and then support numerically. Use limit notation.

a) f(x) = -x3 + x2 + 4x - 4Answer: lim f(x) = ∞ lim f(x) = -

∞ x -∞ x ∞

b) f(x) = (3x – 2)/(x + 1)Answer: both limits are 3

USIN’ YOUR NOGGINTry to determine the end behavior or limit of each function as x approaches infinity without a calculator.

a) f(x) = 1/x2 the limit is o

b) f(x) = 6x – 1 3x + 2the limit is 2

c) f(x) = 2x3/(x – 1)the limit is ∞