Pauls Online Notes : Calculus I

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Online Notes / Calculus I / = Limits /=20 Continuity

Calculus I

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Limits At Infinity, Part II =	E-Book<= /A>=20 Chapte= r=20 Secti= on=20	Th= e=20 Definition of the Limit =

Continuity

Over the last few sections = we=E2=80=99ve been using the=20 term =E2=80=9Cnice enough=E2=80=9D to define those functions = that we could evaluate=20 limits by just evaluating the function at the point in=20 question. It=E2=80=99s now time to formally define = what we mean by=20 =E2=80=9Cnice enough=E2=80=9D.

Definition

A function is said to be continuous at if

= ; = &= nbsp; &n= bsp; &nb= sp; =20

A function is said to be = continuous on the=20 interval [a, b] if it is continuous at each = point in=20 the interval.

This definition can be turned around = into the=20 following fact.

Fact 1

If is continuous at then,

= ; =20

This is exactly the same fact that we = first put=20 down back=20 when we started looking at limits with the exception that we = have=20 replaced the phrase =E2=80=9Cnice enough=E2=80=9D with = continuous.

It=E2=80=99s nice to finally know what = we mean by =E2=80=9Cnice=20 enough=E2=80=9D, however, the definition doesn=E2=80=99t = really tell us just what it=20 means for a function to be continuous. Let=E2=80=99s = take a look at an=20 example to help us understand just what it means for a = function to=20 be continuous.

Example 1=20 Given the graph of f(x), shown = below,=20 determine if f(x) is continuous at , , and .

3D"Continuity_G1"

Solution

To answer the question for each = point we=E2=80=99ll=20 need to get both the limit at that point and the = function=20 value at that point. If they are equal the = function is=20 continuous at that point and if they aren=E2=80=99t = equal the function=20 isn=E2=80=99t continuous at that point.

First .

= ; = &= nbsp; =20

The function value and the limit = aren=E2=80=99t the=20 same and so the function is not continuous at this=20 point. This kind of discontinuity in a graph is = called a=20 jump discontinuity. Jump discontinuities = occur=20 where the graph has a break in it is as this graph = does.

Now .

= ; = &= nbsp; =20

The function is continuous at = this point=20 since the function and limit have the same value.

Finally .

= ; = &= nbsp; =20

The function is not continuous at = this=20 point. This kind of discontinuity is called a=20 removable discontinuity. Removable=20 discontinuities are those where there is a hole in the = graph=20 as there is in this = case.

From this example we can get a quick = =E2=80=9Cworking=E2=80=9D=20 definition of continuity. A function is continuous on = an=20 interval if we can draw the graph from start to finish = without ever=20 once picking up our pencil. The graph in the last = example has=20 only two discontinuities since there are only two places = where we=20 would have to pick up our pencil in sketching it.

In other words, a function is = continuous if its=20 graph has no holes or breaks in it.

For many functions it=E2=80=99s easy to = determine where=20 it won=E2=80=99t be continuous. Functions = won=E2=80=99t be continuous where we=20 have things like division by zero or logarithms of = zero. Let=E2=80=99s=20 take a quick look at an example of determining where a = function is=20 not continuous.

Example 2=20 Determine where the function below is = not=20 continuous.

= ; = &= nbsp; &n= bsp; &nb= sp;=20

Solution

Rational functions are continuous = everywhere except where we have division by = zero. So all=20 that we need to is determine where the denominator is=20 zero. That=E2=80=99s easy enough to determine by = setting the=20 denominator equal to zero and solving.

= ; = &= nbsp; &n= bsp; =20

So, the function will not be = continuous at=20 t=3D-3 and = t=3D5.

A nice consequence of continuity is the = following=20 fact.

Fact 2

If is continuous at and then,

= ; = &= nbsp; &n= bsp; =20

To see a proof of this fact see the Proof=20 of Various Limit Properties section in the Extras = chapter. =20 With this fact we can now do limits like the following = example.

Example 3=20 Evaluate the following limit.

= ; = &= nbsp; &n= bsp; &nb= sp; =20

Solution

Since we know that exponentials = are=20 continuous everywhere we can use the fact above.

= ; = &= nbsp; &n= bsp; =20

Another very nice consequence of = continuity is=20 the Intermediate Value Theorem.

Intermediate Value Theorem

Suppose that f(x) is = continuous on=20 [a, b] and let M be any number between=20 f(a) and f(b). Then there = exists a=20 number c such that,

All the Intermediate Value Theorem is = really=20 saying is that a continuous function will take on all values = between=20 f(a) and f(b). Below is a graph of a = continuous=20 function that illustrates the Intermediate Value = Theorem.

3D"Continuity_G2"

As we can see from this image if we = pick any=20 value, M, that is between the value of f(a) = and the=20 value of f(b) and draw a line straight out from this = point=20 the line will hit the graph in at least one point. In = other=20 words somewhere between a and b the function = will take=20 on the value of M. Also, as the figure shows = the=20 function may take on the value at more than one place.

It=E2=80=99s also important to note = that the Intermediate=20 Value Theorem only says that the function will take on the = value of=20 M somewhere between a and b. It = doesn=E2=80=99t=20 say just what that value will be. It only says that it = exists.

So, the Intermediate Value Theorem = tells us that=20 a function will take the value of M somewhere between = a and b but it doesn=E2=80=99t tell us where = it will take the=20 value nor does it tell us how many times it will take the=20 value. There are important idea to remember about the=20 Intermediate Value Theorem.

A nice use of the Intermediate Value = Theorem is=20 to prove the existence of roots of equations as the = following=20 example shows.

Example 4=20 Show that has a root somewhere in the interval=20

[-1,2].

Solution

What we=E2=80=99re really asking = here is whether or=20 not the function will take on the value

= ; = &= nbsp; &n= bsp; &nb= sp; =20

somewhere between -1 and 2. = In other=20 words, we want to show that there is a number c = such=20 that and . However if we define and acknowledge that and we can see that these two condition on=20 c are exactly the conclusions of the = Intermediate Value=20 Theorem.

So, this problem is set up to use = the=20 Intermediate Value Theorem and in fact, all we need to = do is=20 to show that the function is continuous and that is between and (i.e. or and we=E2=80=99ll be done.

To do this all we need to do is=20 compute,

= ; = &= nbsp; =20

So we have,

= ; = &= nbsp; &n= bsp; =20

Therefore is between and and since is a polynomial it=E2=80=99s continuous=20 everywhere and so in particular it=E2=80=99s = continuous on the=20 interval [-1,2]. So by the Intermediate Value = Theorem=20 there must be a number so that,

= ; = &= nbsp; &n= bsp; &nb= sp; =20

Therefore the polynomial does = have a root=20 between -1 and 2.

For the sake of completeness here = is a=20 graph showing the root that we just proved = existed. Note=20 that we used a computer program to actually find the = root and=20 that the Intermediate Value Theorem did not tell us = what this=20 value was.

3D"Continuity_G3"

Let=E2=80=99s take a look at another = example of the=20 Intermediate Value Theorem.

Example 5=20 If possible, determine if takes the following values in the=20 interval [0,5].

(a) = Does ? [Solution]

(b) = Does ? [Solution]

Solution

Okay, so much as the previous = example we=E2=80=99re=20 being asked to determine, if possible, if the function = takes=20 on either of the two values above in the interval = [0,5]. =20 First, let=E2=80=99s notice that this is a continuous = function and so=20 we know that we can use the Intermediate Value Theorem = to do=20 this problem.

Now, for each part we will let = M be=20 the given value for that part and then we=E2=80=99ll = need to show that=20 M lives between and . If it does then we can use the=20 Intermediate Value Theorem to prove that the function = will=20 take the given value.

So, since we=E2=80=99ll need the = two function=20 evaluations for each part let=E2=80=99s give them = here,

&= nbsp; &n= bsp; &nb= sp; =20

Now, let=E2=80=99s take a look at = each part.

(a)=20 Okay, in this case we=E2=80=99ll define and we can see that,

&= nbsp; &n= bsp; &nb= sp; =20

So, by the Intermediate Value = Theorem there=20 must be a number such that

&= nbsp; &n= bsp; &nb= sp; &nbs= p; = ; =20

[Return to = Problems]

(b) In=20 this part we=E2=80=99ll define . We now have a problem. In this=20 part M does not live between and . So, what does this mean for us? =20 Does this mean that in [0,5]?

Unfortunately for us, this = doesn=E2=80=99t mean=20 anything. It is possible that in [0,5], but is it also possible that in [0,5]. The Intermediate Value=20 Theorem will only tell us that c=E2=80=99s will = exist. =20 The theorem will NOT tell us that c=E2=80=99s = don=E2=80=99t exist.

In this case it is not possible = to=20 determine if in [0,5] using the Intermediate Value=20 Theorem.

[Return to = Problems]

Okay, as the previous example has = shown, the=20 Intermediate Value Theorem will not always be able to tell = us what=20 we want to know. Sometimes we can use it to verify = that a=20 function will take some value in a given interval and in = other cases=20 we won=E2=80=99t be able to use it.

For completeness sake here is the graph = of in the interval [0,5].

3D"Continuity_G4"

From this graph we can see that not = only does in [0,5] it does so a total of 4 times! =20 Also note that as we verified in the first part of the = previous=20 example in [0,5] and in fact it does so a total of 3=20 times.

So, remember that the Intermediate = Value Theorem=20 will only verify that a function will take on a given = value. =20 It will never exclude a value from being taken by the=20 function. Also, if we can use the Intermediate Value = Theorem=20 to verify that a function will take on a value it never = tells us how=20 many times the function will the value, it only tells us = that it=20 does take the value.

Limits At Infinity, Part II =

Th= e=20 Definition of the Limit =

Online Notes / Calculus I / = Limits /=20 Continuity

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