math1401 intro to limits and continuity
TRANSCRIPT
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MATH 1401
SETON HALL UNIVERSITY
Introduction toLIMITS AND CONTINUITY
MATH 1401
PROF. MICHAEL ARYEE 2 SETON HALL UNIVERSITY
Consider the graphs of the two function below.
.1,1
1)(and1)(
32
x
x
xxgxxxf
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MATH 1401
PROF. MICHAEL ARYEE 5 SETON HALL UNIVERSITY
The limits technique provides us with the
tools we need in order to predictthe value or
height of the function by using the values of
the independent variable nearsome given
point, while completely ignoring the value of
the function at that point.
MATH 1401
PROF. MICHAEL ARYEE 6 SETON HALL UNIVERSITY
So, how can we estimate (or guess) the value of thefunction
when x = 1?
First, we can investigate the behavior of the graph ofg
nearx= 1 by using the two sets of values of x:
(1) one set that approaches 1 from theleftof x = 1
x-Values From thelefttowards 1: {0, 0.5, 0.9, 0.99, 0.999}
(2) one set that approaches 1 from therightof x = 1
x-Values From therighttowards 1:{2, 1.5, 1.1, 1.01, 1.001}
1,1
1)(
3
x
x
xxg
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MATH 1401
PROF. MICHAEL ARYEE 7 SETON HALL UNIVERSITY
The answer can be found by numerically investigating the
behavior of the graph ofgas x gets closer and closer to 1 fromthe left and from the right.
Does g(x) appear to get closer and closer to a fixed number?
Question: What
happens to g(x) asx gets closer to 1?
MATH 1401
PROF. MICHAEL ARYEE 8 SETON HALL UNIVERSITY
From the table, we see that when x is close to 1(that is, on either side of 1), g(x) is close to 3.
In fact, it appears that we can make the values ofg(x) asclose as we like to 3 by taking xsufficiently close to 1.For example we can consider x = 0.9999 and 1.0001
We can therefore express the above investigation by
saying that the limit of the function g(x) as x
approaches 1 is equal to 3.
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MATH 1401
PROF. MICHAEL ARYEE 9 SETON HALL UNIVERSITY
The above example shows how to estimate thevalue of a function numerically using the concept oflimits.
The notation for the above example is:
We will introduce the informal definition of limits andthen show more examples of the above.
limx
x
x
1
31
13
MATH 1401
PROF. MICHAEL ARYEE 10 SETON HALL UNIVERSITY
Consider a function fwith an independent variable
xdefined on an open interval (a, b), except for somefixed numberc, which also belongs to the interval,
then the function fis said to approach L(the limit),as xapproaches c, if the value of f(x) get closer and
closer to a unique number L as values of x getscloser and closer to c.
Using symbols we write:
lim ( )x c
f x L
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MATH 1401
PROF. MICHAEL ARYEE 11 SETON HALL UNIVERSITY
Note that cand L are both numbers and they satisfy
the following conditions:
Wheneverxis close to c, but not equal to c, f(x) isclose to L.
As x gets closer and closer to c, but not equal to c,f(x) gets closer and closer to L.
We can make f(x) close to L as we please bychoosing xto be close to cbut not equal to c.
We want the difference between f(x) and L to be asnear zero as possible for all values ofxdefined onthe open interval (a, b) containing c.
MATH 1401
PROF. MICHAEL ARYEE 12 SETON HALL UNIVERSITY
When we consider the limit from only one direction,
either from the left only, or from the right only, the
answer is regarded as a one-sided limit.
A one-sided limit can be a Left-hand limit or aRight-hand limit.
We denote a Left-Hand Limit with a little negativesymbol (-) at the location where the exponentof thenumber should be. We write:
We denote a Right-Hand Limit with a little positivesymbol (+) at the location where the exponentof thenumber should be. We write:
lim ( )x c
f x
lim ( )x c
f x
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MATH 1401
PROF. MICHAEL ARYEE 13 SETON HALL UNIVERSITY
Using our table below:
Left Handg x
x
:lim ( )
1
3Right Hand
g xx
:
lim ( )1
3
MATH 1401
PROF. MICHAEL ARYEE 14 SETON HALL UNIVERSITY
In general, the limit of f(x) existsifallof the followingstatements are true:
1. is definite and has a value.
2. is definite and has a value.
3 =
lim ( )x c
f x
lim ( )x c
f x
lim ( )
x cf x
lim ( )x c
f x
if and only if andlim ( )x c
f x L
lim ( )
x cf x L
lim ( )x c
f x L
In General:
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MATH 1401
PROF. MICHAEL ARYEE 15 SETON HALL UNIVERSITY
1.
2.
3.
lim ( )x
g x
1
3
lim ( )x g x
1 lim ( )x g x
1 3
lim ( )x
g x
1
3
if and only if andlim ( )x
g x
1
3lim ( )x
g x
1
3 lim ( )x
g x
1
3
That is:
MATH 1401
PROF. MICHAEL ARYEE 16 SETON HALL UNIVERSITY
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MATH 1401
PROF. MICHAEL ARYEE 17 SETON HALL UNIVERSITY
MATH 1401
PROF. MICHAEL ARYEE 18 SETON HALL UNIVERSITY
Lets consider the function whose graph is
represented below.
If we inspect the graph, we can see that there is a
breakat x= 4.
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MATH 1401
PROF. MICHAEL ARYEE 19 SETON HALL UNIVERSITY
If we travel towards x= 4 from the right, we will
arrive at the height of 2.
.
The height at which we arrive at if we are traveling
from the RIGHT is called the Right-Hand Limit of f(x)as xapproaches a specific value ofx, in this case asxapproaches 4.
MATH 1401
PROF. MICHAEL ARYEE 20 SETON HALL UNIVERSITY
If we travel towards x= 4 from the left, we will arriveat the height of 1.
The height at which we arrive at if we are traveling
from the LEFT is called the Left-Hand Limit off(x)as xapproaches a specific value ofx, in this case asxapproaches 1.
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MATH 1401
PROF. MICHAEL ARYEE 21 SETON HALL UNIVERSITY
D.N.E. means Does Not Exist.
MATH 1401
PROF. MICHAEL ARYEE 22 SETON HALL UNIVERSITY
To estimate the limit, there are three approaches. We
have just illustrated two of these approaches
(numerical approach and graphical approach). The
three approaches are listed below:
Numerical approach (limit obtained by weconstructing a table of values)
Graphical approach (limit obtained by sketching agraph)
Analytic approach (limit obtained by usingalgebra)
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MATH 1401
PROF. MICHAEL ARYEE 23 SETON HALL UNIVERSITY
Find by means of a table of values.
limx
xx
1
2 1
1
MATH 1401
PROF. MICHAEL ARYEE 24 SETON HALL UNIVERSITY
Find by means of a table of values.
limx
x
x
1
21
1
From the table:
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MATH 1401
PROF. MICHAEL ARYEE 25 SETON HALL UNIVERSITY
Find by means of a table of values.
Alternatively, you can setup a horizontal table like the
one below and determine the limits from both sides.
limx
x
x
1
2 1
1
x 0 0.5 0.9 0.99 0.999 1 1.001 1.01 1.1 1.5 2
1 1.5 1.9 1.99 1.999 2.001 2.01 2.1 2.5 3
limx
x
x
1
21
12
MATH 1401
PROF. MICHAEL ARYEE 26 SETON HALL UNIVERSITY
Given the graph off(x) shown below, Determine thelimit at x = 0.
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MATH 1401
PROF. MICHAEL ARYEE 27 SETON HALL UNIVERSITY
Given the graph off(x) shown below, Determine thelimit at x = 1.
MATH 1401
PROF. MICHAEL ARYEE 28 SETON HALL UNIVERSITY
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MATH 1401
PROF. MICHAEL ARYEE 29 SETON HALL UNIVERSITY
This involves using algebra to find the limit
To find the limit of a function using the analytical
approach, first try direct substitution.
We usually use direct substitution when the function
you are dealing with is a polynomial, a rational, or
any algebraic function whose domain do not exclude
the specific value that x approaches.
Direct substitution is always valid for polynomials and
rational functions with nonzero denominators.
MATH 1401
PROF. MICHAEL ARYEE 30 SETON HALL UNIVERSITY
Iffis a polynomial function, then
Ex.
Ifris a rational function given by
and then
Ex.
lim ( ) ( )x a
p x p a
1-(1)-2(1)3(1)1x2x3xlim2424
1
x
0)( ag
r x f xg x
( ) ( )( )
lim ( ) ( )x a
r x r a
lim( ) ( )
( )x
x x
x
2
2 22
2
2 2 2
2 2
4
41
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MATH 1401
PROF. MICHAEL ARYEE 31 SETON HALL UNIVERSITY
MATH 1401
PROF. MICHAEL ARYEE 32 SETON HALL UNIVERSITY
If the limit of f(x) as x approaches c cannot beevaluated through direct substitution, because thiswill cause denominator to be zero, then try tofactorand cancel out the common factors, and thenusedirect substitution again.
, Note that Direct Substitution =
Now factor, cancel, and do direct substitution again:
0
0lim
x
x x
x
3
2 6
3
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MATH 1401
PROF. MICHAEL ARYEE 33 SETON HALL UNIVERSITY
MATH 1401
PROF. MICHAEL ARYEE 34 SETON HALL UNIVERSITY
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MATH 1401
PROF. MICHAEL ARYEE 35 SETON HALL UNIVERSITY
If the numerator approaches 0 and the denominatoralso approaches 0 and the numerator involves asquare root expression, thenrationalize thenumerator by multiplying both the numerator and
the denominator by the conjugate of the
numerator.
In general if f(x) = g(x) for every x except possibly x =c, in some interval containing c, then
lim ( ) lim ( ) ( )x a x a
f x g x g a
MATH 1401
PROF. MICHAEL ARYEE 36 SETON HALL UNIVERSITY
Find
Multiply numerator and denominator by
conjugate .
limx
x
x
4
2
4
2x
.
4
1
22
1
24
1
2
1lim
4
xx
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MATH 1401
PROF. MICHAEL ARYEE 37 SETON HALL UNIVERSITY
Find
.
MATH 1401
PROF. MICHAEL ARYEE 38 SETON HALL UNIVERSITY
.
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MATH 1401
PROF. MICHAEL ARYEE 39 SETON HALL UNIVERSITY
Given the function
determine whether or not the limit existat x=1.
.
lim ( ) lim ( )x x
f x x
1 1
3 3 1 3
lim ( ) . .x
f x D N E1
Left:
Right: lim ( ) lim ( )x xf x x 1 1 2 3 2 1 3 5
MATH 1401
PROF. MICHAEL ARYEE 40 SETON HALL UNIVERSITY
Determine whether or not the function f(x) = | x 2 |
has a limit at x = 2.
The function f(x)= | x 2 | is the same as:
.
lim ( ) lim[ ( )] ( )x x
f x x
2 2
2 2 2 0
lim ( )x
f x2
0
Left:
Right: lim ( ) lim( ) ( )x x
f x x
2 2
2 2 2 0
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MATH 1401
PROF. MICHAEL ARYEE 41 SETON HALL UNIVERSITY
.
MATH 1401
PROF. MICHAEL ARYEE 42 SETON HALL UNIVERSITY
.
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MATH 1401
PROF. MICHAEL ARYEE 43 SETON HALL UNIVERSITY
.
MATH 1401
PROF. MICHAEL ARYEE 44 SETON HALL UNIVERSITY
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MATH 1401
PROF. MICHAEL ARYEE 45 SETON HALL UNIVERSITY
.
MATH 1401
PROF. MICHAEL ARYEE 46 SETON HALL UNIVERSITY
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PROF. MICHAEL ARYEE 47 SETON HALL UNIVERSITY
.
MATH 1401
SETON HALL UNIVERSITY
CONTINUITY
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MATH 1401
PROF. MICHAEL ARYEE 49 SETON HALL UNIVERSITY
A function fis continuous at a point x = cif thegraph of fhas no gap, break, split, holes, jumps,or a missing point for f(x) at the point x = c.
In addition, a function fis continuous at a point x = cif you can move along the graph offthrough thepoint (c, f(c)) with a pencil without lifting the pencilfrom the paper.
MATH 1401
PROF. MICHAEL ARYEE 50 SETON HALL UNIVERSITY
A function f is continuous at a point x = c if all of the
following THREE conditions are fulfilled:
1. f(c ) is defined
2. exists, and
3. = f(c ).
For a function to be continuous at the point x = c, the functionmust have a value at x = c(which isf(c)) and the limit offas xapproaches c( ) must have a value which is the sameasthe value of the function at x = c
lim ( )x c
f x
lim ( )x c
f x
lim ( )x c
f x
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MATH 1401
PROF. MICHAEL ARYEE 51 SETON HALL UNIVERSITY
A function is said to be discontinuousat x = c, if ahole, gap, or break occurs in the graph at x = c,meaning the function violates one of the three
items above. Below are some examples of
discontinuities at x = c.
MATH 1401
PROF. MICHAEL ARYEE 52 SETON HALL UNIVERSITY
When a function is not defined at certain values,these values are certainly the points of discontinuityof the function.
Because division by zero is not allowed in
mathematics, any number or numbers that when
substituted into the function causes thedenominator to be zero, thus making the function
undefined, represent points of discontinuity.
There are two main categories of discontinuity:
removable and non-removable.
Non-removable has two divisions: a jumpdiscontinuity and an infinite discontinuity.
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MATH 1401
PROF. MICHAEL ARYEE 53 SETON HALL UNIVERSITY
A jump discontinuity
A non-removable jump discontinuity is thesituation where we are unableto find a value for
because
lim ( )x c
f x
lim ( ) lim ( )x c x c
f x f x
MATH 1401
PROF. MICHAEL ARYEE 54 SETON HALL UNIVERSITY
An infinite discontinuity
A non-removable infinite discontinuity is thesituation where we are unableto find a value for
because = infinity.
lim ( )x c
f x
lim ( )x c
f x
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MATH 1401
PROF. MICHAEL ARYEE 55 SETON HALL UNIVERSITY
A removable discontinuity is the situation where
#1) We are able to find a value for , but we are
unableto find a value forf(c),
or
#2) we are able to get values for
both and f(c), however,
their values do not match,
that is, lim ( ) ( )x c
f x f c
lim ( )x c
f x
lim ( )x c
f x
MATH 1401
PROF. MICHAEL ARYEE 56 SETON HALL UNIVERSITY
Given the function
Determine if the function is continuous at x=1.
1. Find f(c):
2. Find 2a.
2b.
3. Compare #1. and #2.
Therefore, the function is continuous at x =1.
.
lim ( ) lim ( )x x
f x x
1 1
3 3 1 3
lim ( )x
f x1
3
Left:
Right: lim ( ) lim ( )x x
f x x
1 1
2 1 2 1 1 3
f( ) ( )1 3 1 3
lim ( )x f x1
lim ( ) ( )x
f x f
1
1 3
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MATH 1401
PROF. MICHAEL ARYEE 57 SETON HALL UNIVERSITY
Given the function
determine if the function is continuous at x=1.
1. Find f(c):
2. Find
2a.
2b.
3. Compare #1. and #2.
Therefore, the function is DISCONTINUOUS at x=1.
.
lim ( ) lim ( )x x
f x x
1 1
3 1 3 1 1 4
lim ( )x
f x DNE1
Left:
Right: lim ( ) lim ( )x x
f x x
1 1
2 3 2 1 3 5
f( ) ( )1 3 1 1 4 lim ( )x
f x1
lim ( ) ( )x
f x f
1
1
MATH 1401
PROF. MICHAEL ARYEE 58 SETON HALL UNIVERSITY
Given the function
Determine if the function is continuous at x=-3.
1. Find f(c):
2. Find
3. Compare #1. and #2.
Therefore, the function is CONTINUOUS at x=-3.
.
f( ) 3 5
lim ( ) ( )x
f x f
3
3 5
lim ( )x f x 3
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MATH 1401
PROF. MICHAEL ARYEE 59 SETON HALL UNIVERSITY
Given the function
Determine the value of a that makes f(x) continuous
We need to find the value of a so that :
(continue on next slide)
.
MATH 1401
PROF. MICHAEL ARYEE 60 SETON HALL UNIVERSITY
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PROF. MICHAEL ARYEE 62 SETON HALL UNIVERSITY
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PROF. MICHAEL ARYEE 64 SETON HALL UNIVERSITY
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PROF. MICHAEL ARYEE 66 SETON HALL UNIVERSITY
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PROF. MICHAEL ARYEE 70 SETON HALL UNIVERSITY
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PROF. MICHAEL ARYEE 71 SETON HALL UNIVERSITY
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MATH 1401.