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

Determining Limit Values

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 - - - - - - - - - - - - - - - - - - - - - - - - - -

- Limits give us a language for describing how the outputs (y values) of a function behave as the inputs (xvalues) approachsome particular value.
- Sometimes we use direct substitution, factoring, rationalizing or expanding to calculate a limit.

We write limf(x) = L

x→a

and say “the limit of f(x), as xapproachesa, equals L”

if we can make the values off(x) arbitrarily close to L

(as close to Las we like)

by taking xto be sufficiently close to a, but not equal to a.

Functions Continuous at ‘a’

A function is continuous when its graph is a single

unbroken curve that you could draw without lifting

your pen from the paper.

That is not a formal definition, but it helps you

understand the idea.

Functions Continuous at ‘a’

Polynomial Functions are Continuousat all x values

(1, 5)

(1 3)

5

3

lim5 =

x→1

5

lim (– 2x + 5) =

x→3

3

Functions Continuous at ‘a’

lim() =

x→3

Functions Continuous at ‘a’

0

0

limf(x) =

x→2

(2, 0)

If direct substitution can be done without an error

we can determine the function value and limit this way.

lim(7 – 2x) =

x→3

7 – 2(3) = 1

7 – 2(3) = 1

42 + 2(4) = 24

lim(x2 + 2x) =

x→4

42 + 2(4) = 24

lim(2x2 – 3x + 4) =

x→2

2(2)2– 3(2) + 4 = 6

2(2)2– 3(2) + 4 = 6

If ris a rational function given by

and c is a real number such that q(c) ≠ 0, then

Let’s look at some examples to clarify this MATH LANGUAGE

Remember: Division by 0 is undefined.

Rational Functions are continuous at any x value that does not result in division by zero.

By Substitution

The function is continuous at x = 3

So what is not continuous

(also called discontinuous)?

Look out for holes, jumps or vertical asymptotes.

Exploring a Nonexistent Limit.

Rational Functions are discontinuous at any x value that DOES result in division by zero.

undefined

There will be a vertical asymptote

when both function and limit

do not exist

Exploring a Nonexistent Limit.

x = 1

Use a table to show the

limit does not exist.

Find

- Use a graph to show the
- limit does not exist.

The y values are getting further

apart the closer x gets to 1

Since f (1) does not exist and the limit as x→1 does not exist

there is a vertical asymptote at x = 1

Direct substitution

won’t WORK!!!

DNE

Factor and reduce

There is a vertical asymptote

f(5) does not exist

Substitute – 5 in forx

Undefined!

Direct substitution

won’t WORK!!!

undefined

HOWEVER as x→5

Factor and reduce

f(5) does not exist

There is a “hole” at

Substitute 5 in for x

-∞

Approaching 0.1666..

∞

Lesson #5 Worksheet EXAMPLE 1: Continued

The function f(x) does not exist at x= +3 but the limits of f(x) as x approaches are a different matter.

(-3, 0.166666)

x = 3

undefined

The function g has a limit of 2 as x →1

even though g(1) ≠ 2

(1, 2)

(1, 1)

(1, 1)

The function is discontinuous when x = 1.

There is a “hole” at (1, 2).

YES!

Is there a radical rule?

Let n be a positive integer. The following limit is valid

for all c if n is odd, and is valid for c > 0 if n is even.

n is odd and > 0

.44

By Graphing

By Substitution

(3, 2)

2

3

Use a graph of to show that the following limit does not exist.

There is no point on

the graph wherex= – 3

By substitution:

is a non-real number

Multiply by its conjugate

3

NOTE: The answer is a rational number

Rational number

Direct substitution won’t WORK!!!

1

Direct substitution won’t WORK!!!

Multiply Top and Bottom by the conjugate.

1

1

Again substitution won’t WORK.

Find