- 77 Views
- Uploaded on
- Presentation posted in: General

Section 1.4 – Continuity and One-Sided Limits

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Section 1.4 – Continuity and One-Sided Limits

When x=5, all three pieces must have a limit of 8.

Find values of a and b that makes f(x) continuous.

f(x)

L

x

c

For every question of this type, you need (1), (2), (3), conclusion.

A function f is continuous at c if the following three conditions are met:

- is defined.
- exists.

The function is clearly defined at x= 0

With direct substitution the limit clearly exists at x=0

The value of the function clearly equals the limit at x=0

f is continuous at x = 0

Show is continuous at x = 0.

The function is clearly 10 at x = 2

With direct substitution the limit clearly exists at x=0

The behavior as x approaches 2 is dictated by 8x-1

The value of the function clearly does not equal the limit at x=2

f is not continuous at x = 2

Show is not continuous at x = 2.

Typically a hole in the curve

Step/Gap

Asymptote

If f is not continuous at a, we say f is discontinuous at a, or f has a discontinuity at a.

Find the x-value(s) at which is not continuous. Which of the discontinuities are removable?

If f can be reduced, then the discontinuity is removable:

Notice that:

This is the same function as f except at x=-3

There is a discontinuity at x=-3 because this makes the denominator zero.

f has a removable discontinuity at x = -3

Let:

f(x)

L

If:

x

c

Then f(x) has a removable discontinuity at x=c.

If f(x) has a removable discontinuity at x=c.

Then the limit of f(x) atx=c exists.

If f(x) becomes arbitrarily close to a single REAL number L as x approaches c from values less than c, the left-hand limit is L.

f(x)

L

x

c

The limit of f(x)…

is L.

Notation:

as x approaches c from the left…

If f(x) becomes arbitrarily close to a single REAL number L as x approaches c from values greater than c, the right-hand limit is L.

f(x)

L

x

c

The limit of f(x)…

is L.

Notation:

as x approaches c from the right…

Evaluate the following limits for

Analytically find .

f(x)

L

x

c

A limit exists if…

Left-Hand Limit

=

Right-Hand Limit

Let fbe a function and let c be real numbers. The limit of f(x)as x approaches c is Lif and only if

Use when x>2

Use when x<2

You must use the piecewise equation:

Analytically show that .

You must use the piecewise equation:

Use when x>-1

Use when x<-1

Analytically show that is continuous at x = -1.

f(b)

f(a)

x

a

b

Must have closed dots on the endpoints.

A function f is continuous on [a, b]if it is continuous on (a, b) and

Use the graph of t(x) to determine the intervals on which the function is continuous.

Are the one-sided limits of the endpoints equal to the functional value?

By direct substitution:

The domain of f is [-1,1]. From our limit properties, we can say it is continuous on (-1,1)

f is continuous on [-1,1]

Discuss the continuity of

Is the middle is continuous?

- Scalar Multiple:
- Sum/Difference:
- Product:
- Quotient: if
- Composition:
- Example: Since are continuous, is continuous too.

If b is a real number and f and g are continuous at x = c, then following functions are also continuous at c:

This theorem does NOT find the value of c. It just proves it exists.

f(b)

k

f(a)

c

a

b

If f is continuous on the closed interval [a, b] and k is any number between f(a) and f(b), then there is at least one number c in [a, b] such that:

Notice how every part of the theorem is discussed (values of the function AND continuity).

We will learn later that this implies continuity.

Since h(3) < -5 < h(1) and h is continuous, by the IVT, there exists a value r, 1 < r < 3, such that h(r) = -5.

Find an output less than zero

Find an output greater than zero

Since f(0) < 0 and f(2) > 0

There must be some csuch that f(c) = 0 by the IVT

The IVT can be used sincef is continuous on [-∞,∞].

Use the intermediate value theorem to show has at least one root.

Solve the equation for zero.

Find an output greater than zero

Find an output less than zero

Since and

The IVT can be used sincethe left and right side are both continuous on [-∞,∞].

There must be some csuch that cos(c) = c3 - c by the IVT

Show that has at least one solution on the interval .