# MathematicsClick Here For Details on How to Complete this Paper……Need a Professional Writer to Work on this Paper and Give you a 100 % Original Paper? Click Here and Get a Professional Writer to work on this Essay…… 1. Direct replacement in limit problemsWe noted that when we solved limit problems limx?a f(x), sometimesthe final answer was just f(a).Example.limx?3??x2 + x + 1 # \$% & f(x)?? = limx?3x2 + limx?3x + limx?31=)limx?3x*2+3+1=32 +3+1 # \$% & f(a).We could have gotten the final answer by replacing x with a. In otherwords the behavior near x = a, limx?a f(x) was determined by the valuef(a). We can actually do this for a large class of functions.23A function continuous atx = 2, f(2) is defined, andlimx?2 f(x) exists, andlimx?2 f(x)=3= f(2).2. Continuity – behavior near a number is determined by thevalue at the number23A function not continuousat x = 2, since f(2) is notdefined.231A function not continuousat x = 2. Why? sincef(2) = 1 butlimx?2 f(x)=3.Definition. A function f(x) is continuous at x = a ifIn practice the graphs of functions that are continuous at x = a do nothave gaps or jumps at x = a. They usually can be sketched without liftingoff the pencil. However, there are exceptions (Problem 11).23An example of a function discontinuous (not continuous) at x = 2 sincelimx?2 f(x) does not exist. (the left and right limits as x ? 2 are different:limx?2- f(x) = 3 but limx?2+ f(x) = 1).3. Removable discontinuitiesIf limx?a f(x) exists but limx?a f(x) ?= f(a), then f(x) has a removablediscontinuity at x = a. If we redefine f(a) to bef(a) = limx?a f(x), then the new function is continuous at x = a. 12For example,f(x) = + x + 1 if x ?= 21.5 if x = 2has a discontinuity at x = 2 since limx?2 f(x) = 3 but f(2) = 1.5. Thediscontinuity x = 1.5 can be removed by redefining f(2) = 3.4. Jump discontinuityA jump discontinuity occurs at a number x = a if both limits limx?a- f(x)and limx?a+ f(x) exist but they are different.21.53An example of a jumpdiscontinuity.limx?2- f(x)=3 butlimx?2+ f(x)=1.5.5. Classes of functions that are continuousThe first example in this section is a polynomial. In general,Theorem 1 (Polynomials are continuous). Polynomials (i.e. f(x) = x3 –x2 + 4 ) are functions continuous everywhere.We prove this theorem using the definition of continuity and the LimitsLaws.Let f(x) = c0 + c1x + ··· + cnxn. Then,limx?af(x) = limx?a(c0 + c1x + ··· + cnxn (1) )= limx?ac0 + limx?ac1x + ··· + limx?acnxn (2)= c0 + c1 limx?ax + ··· + cn limx?axn (3)= c0 + c1a + ··· + cnan (4) = f(a).Without proof we state that alsoTheorem 2. Rational functions (i.e f(x) = x3-x2+4x2-x-4 ), exponential functions(i.e. f(x) = ex), logarithmic functions (i.e. f(x) = ln x), rootfunctions ((i.e. f(x) = v3x) are all continous whenever they are defined.Theorem 3. An inverse function of a continuous function is continuous.Theorem 4. Also if f and g are continuous then f + g, f – g, f · g andf/g are continuous functions whenever they are defined.For example, let f(x) = x3 – x2 + 4 and g(x) = ex. Then both f and gare continuous. Also x3 – x2 +4+ ex is continous, so are (x3 – x2 + 4)exand x3-x2+4ex .36. CompositionThe limit of a continuous function can be computed by evaluating thefunction at the limit. In other words the order of applying the limit andthe function can be changed.Theorem 5 (A continuous function commutes with the limit). Let f becontinuous then,limx?af(g(x)) = f(limx?ag(x)).This property is sometimes used when we solve limit problems. Alsonote that if g is continuous then the right limit is just f(g(a)). So if we letF = f(g(x)) then limx?a F(x) = F(a). We summarize this asTheorem 6 (The composition of continuous functions is continuous). Ifg is continuous at a number a and f is continuous at g(a), then F = f ? gis continuous at a.For example the function F(x) = vx2 + 1 is continuous for all x sinceit is the composition of f(x) = vx and g(x) = x2 + 1, both of which arecontinuous on their domains.We can also now detemine continuity of complicated functions built bycompositions and elementary operations without using the definition of thelimit. For example, we explain that the function f(x) = ln(x)+vx2 + 1·exis continuous in its domain x > 0.Note that vx2 + 1 is continuous so is ex thus their product vx2 + 1 · exis continuous. Since ln(x) is continuous finally we conclude that f(x) iscontinuous.7. Intermediate Value TheoremA key propetry of a continuous function f(x) on a = x = b is that itassumes all values between f(a) and f(b). The continous function f(x)cannot skip a number between f(a) and f(b) (otherwise there would be agap on the graph of f(x)).a c bf(a)Nf(b)Theorem 7 (Intermediate Value Theorem). Let f be continuous on theinterval [a, b]. Then for every value N between f(a) and f(b), there is somec in [a, b] such that f(c) = N.There are functions with a discontinuities that are neither removable norjump discontinuities. Can you sketch one?48. Continuous from one sideaf(a)Continuous from the left. In the figure on the margin, f(x) is not continuous at x = a sincelimx?a f(x) does not exist. However, as the values of x approach a fromthe left, then the values of f(x) approach the value f(a). In this case,Let f(x) be defined on an open interval I. If f is continuous at eachnumber in I then we say that f is continuous on the interval I.For example let f(x) = v1x be defined on (0, 1). Since f is the fractionof two functions, both of which are continuous on the interval (0, 1), andsince the denominator is not zero on (0, 1), f is a continuous function onthe interval (0, 1).There is a subtlety what do if the domain of f is not open. For example,let I = (0, 1]. Then we only evaluate limx?1- f(x) as f is not defined forx > 1. Thus we can only determine whether limx?1- f(x) = f(1), that isif f is continuous from the left at the right endpoint x = 1.We agree then to say that a function f(x) defined on an interval I = (a, b]is continuous on I if f is continuous at each number inside I and if f iscontinuous from the right at x = b.Similarly, we define continuity of intervals [a, b), [a, b], [a, 8), and (-8, b].Click Here For Details on How to Complete this Paper……Need a Professional Writer to Work on this Paper and Give you a 100 % Original Paper? Click Here and Get a Professional Writer to work on this Essay……

Other Services

Mathematics

Need a Professional Writer to Work on this Paper and Give you a 100 % Original Paper? Click Here and Get a Professional Writer to work on this Essay……
1. Direct replacement in limit problems
We noted that when we solved limit problems limx?a f(x), sometimes
the final answer was just f(a).
Example.
limx?3
?
?x2 + x + 1 # \$% & f(x)
?
? = limx?3
x2 + limx?3
x + limx?3
1
=
)
limx?3
x
*2
+3+1=32 +3+1 # \$% & f(a)
.
We could have gotten the final answer by replacing x with a. In other
words the behavior near x = a, limx?a f(x) was determined by the value
f(a). We can actually do this for a large class of functions.
2
3
A function continuous at
x = 2, f(2) is defined, and
limx?2 f(x) exists, and
limx?2 f(x)=3= f(2).
2. Continuity – behavior near a number is determined by the
value at the number
2
3
A function not continuous
at x = 2, since f(2) is not
defined.
2
3
1
A function not continuous
at x = 2. Why? since
f(2) = 1 but
limx?2 f(x)=3.
Definition. A function f(x) is continuous at x = a if
In practice the graphs of functions that are continuous at x = a do not
have gaps or jumps at x = a. They usually can be sketched without lifting
off the pencil. However, there are exceptions (Problem 11).
2
3
An example of a function discontinuous (not continuous) at x = 2 since
limx?2 f(x) does not exist. (the left and right limits as x ? 2 are different:
limx?2- f(x) = 3 but limx?2+ f(x) = 1).
3. Removable discontinuities
If limx?a f(x) exists but limx?a f(x) ?= f(a), then f(x) has a removable
discontinuity at x = a. If we redefine f(a) to be
f(a) = limx?a f(x), then the new function is continuous at x = a. 1
2
For example,
f(x) = + x + 1 if x ?= 2
1.5 if x = 2
has a discontinuity at x = 2 since limx?2 f(x) = 3 but f(2) = 1.5. The
discontinuity x = 1.5 can be removed by redefining f(2) = 3.
4. Jump discontinuity
A jump discontinuity occurs at a number x = a if both limits limx?a- f(x)
and limx?a+ f(x) exist but they are different.
2
1.5
3
An example of a jump
discontinuity.
limx?2- f(x)=3 but
limx?2+ f(x)=1.5.
5. Classes of functions that are continuous
The first example in this section is a polynomial. In general,
Theorem 1 (Polynomials are continuous). Polynomials (i.e. f(x) = x3 –
x2 + 4 ) are functions continuous everywhere.
We prove this theorem using the definition of continuity and the Limits
Laws.
Let f(x) = c0 + c1x + ··· + cnxn. Then,
limx?a
f(x) = limx?a
(c0 + c1x + ··· + cnxn (1) )
= limx?a
c0 + limx?a
c1x + ··· + limx?a
cnxn (2)
= c0 + c1 limx?a
x + ··· + cn limx?a
xn (3)
= c0 + c1a + ··· + cnan (4) = f(a).
Without proof we state that also
Theorem 2. Rational functions (i.e f(x) = x3-x2+4
x2-x-4 ), exponential functions
(i.e. f(x) = ex), logarithmic functions (i.e. f(x) = ln x), root
functions ((i.e. f(x) = v3x) are all continous whenever they are defined.
Theorem 3. An inverse function of a continuous function is continuous.
Theorem 4. Also if f and g are continuous then f + g, f – g, f · g and
f/g are continuous functions whenever they are defined.
For example, let f(x) = x3 – x2 + 4 and g(x) = ex. Then both f and g
are continuous. Also x3 – x2 +4+ ex is continous, so are (x3 – x2 + 4)ex
and x3-x2+4
ex .
3
6. Composition
The limit of a continuous function can be computed by evaluating the
function at the limit. In other words the order of applying the limit and
the function can be changed.
Theorem 5 (A continuous function commutes with the limit). Let f be
continuous then,
limx?a
f(g(x)) = f(limx?a
g(x)).
This property is sometimes used when we solve limit problems. Also
note that if g is continuous then the right limit is just f(g(a)). So if we let
F = f(g(x)) then limx?a F(x) = F(a). We summarize this as
Theorem 6 (The composition of continuous functions is continuous). If
g is continuous at a number a and f is continuous at g(a), then F = f ? g
is continuous at a.
For example the function F(x) = v
x2 + 1 is continuous for all x since
it is the composition of f(x) = vx and g(x) = x2 + 1, both of which are
continuous on their domains.
We can also now detemine continuity of complicated functions built by
compositions and elementary operations without using the definition of the
limit. For example, we explain that the function f(x) = ln(x)+vx2 + 1·ex
is continuous in its domain x > 0.
Note that vx2 + 1 is continuous so is ex thus their product vx2 + 1 · ex
is continuous. Since ln(x) is continuous finally we conclude that f(x) is
continuous.
7. Intermediate Value Theorem
A key propetry of a continuous function f(x) on a = x = b is that it
assumes all values between f(a) and f(b). The continous function f(x)
cannot skip a number between f(a) and f(b) (otherwise there would be a
gap on the graph of f(x)).
a c b
f(a)
N
f(b)
Theorem 7 (Intermediate Value Theorem). Let f be continuous on the
interval [a, b]. Then for every value N between f(a) and f(b), there is some
c in [a, b] such that f(c) = N.
There are functions with a discontinuities that are neither removable nor
jump discontinuities. Can you sketch one?
4
8. Continuous from one side
a
f(a)
Continuous from the left. In the figure on the margin, f(x) is not continuous at x = a since
limx?a f(x) does not exist. However, as the values of x approach a from
the left, then the values of f(x) approach the value f(a). In this case,
Let f(x) be defined on an open interval I. If f is continuous at each
number in I then we say that f is continuous on the interval I.
For example let f(x) = v
1
x be defined on (0, 1). Since f is the fraction
of two functions, both of which are continuous on the interval (0, 1), and
since the denominator is not zero on (0, 1), f is a continuous function on
the interval (0, 1).
There is a subtlety what do if the domain of f is not open. For example,
let I = (0, 1]. Then we only evaluate limx?1- f(x) as f is not defined for
x > 1. Thus we can only determine whether limx?1- f(x) = f(1), that is
if f is continuous from the left at the right endpoint x = 1.
We agree then to say that a function f(x) defined on an interval I = (a, b]
is continuous on I if f is continuous at each number inside I and if f is
continuous from the right at x = b.
Similarly, we define continuity of intervals [a, b), [a, b], [a, 8), and (-8, b].