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2

LIMITS

Section 2:  Theorems on limits

Limits of polynomials

Back to Section 1

We will now prove that a certain limit exists, namely the limit of f (x) = x  as x approaches any value c.  (That f(x) also approaches c should be obvious.)

 

 

THEOREM.   If f (x) = x, then for any value c that we might name:

Theorems on limits

Theorems on limits

For, if a sequence of values of the variable x approaches c as a limit (Definition 2.1), then a sequence of values of the function  f(x) = x will also approach c as a limit (Definition 2.2).

 

For example,

Theorems on limits

Theorems on limits

Theorems on limits

To help us calculate limits, it is possible to prove the following.

 

Let f and g be functions of a variable x.  Then, if the following limits exist:

Theorems on limits

1)   Theorems on limits  (f + g)  =  A + B.
 
2)   Theorems on limits  (f g)  =  AB.
3)   Theorems on limits  f
g
 =   A
B
,  if B is not 0.

In other words:

1)  The limit of a sum is equal to the sum of the limits.

2)  The limit of a product is equal to the product of the limits.

3)  The limit of a quotient is equal to the quotient of the limits,
3)  provided the limit of the denominator is not 0.

Also, if c does not depend on x -- if c is a constant -- then

4)   Theorems on limits

Theorems on limits 

To see that, let x approach 4:  e.g.  4 1
2
  4 1
4
  4 1
8
  4  1 
16
  4  1 
32
  .  .  . , then

the value of 5 -- or any constant -- does not change.  It is constantExclamation

When c is a constant factor, but f depends on x, then

5)   Theorems on limits

A constant factor may pass through the limit sign.  (This follows from Theorems 2 and 4.)  For example,

Theorems on limits

Example 1.   Quote Theorems 1) through 5) to prove the following:

Theorems on limits

  Solution.   x2 = x· x.  And we have proved that  Theorems on limits  exists, and is
  equal to 4.  Therefore,   Theorems on limits  according to Theorem 2.

That is,

Theorems on limits

It should be clear from this example that to evaluate the limit of any power of x  as x approaches any value, simply evaluate the power at that value.  Repeated application of Theorem 2 affirms that.

   Problem 2.    Theorems on limits

Theorems on limits

Problem 3.   Evaluate the following limits, and justify your answers by quoting Theorems 1 through 5.

Theorems on limits 

43 + 4 = 64 + 4 = 68. This follows from Theorem 1 and Theorem 2.

Theorems on limits 

42 + 1 = 16 + 1 = 17. This follows from Theorem 1, Theorem 2, and Theorem 4.

Theorems on limits 

5· 42 = 5· 16 = 80. This follows from Theorem 5 and Theorem 2.

Theorems on limits 

15
 5
 =  3.

The limits of the numerator and denominator follow from Theorems 1, 2, and 4.  The limit of the fraction follows from Theorem 3.

Limits of polynomials

The student might think that to evaluate a limit as x approaches a value, all we do is evaluate the function at that value.  And for the most part that is trueExclamation  One of the most important classes of functions for which that is true are the polynomials.  (Topic 6 of Precalculus.)  A polynomial in x has this general form:

Theorems on limits

where n is a whole number, and annot equals0.

Therefore, according to the Theorems on limits, to name the limit of a polynomial as x approaches any value c, simply evaluate the polynomial at that value.

If P(x) is a polynomial, then

Theorems on limits

Compare Example 1 and Problem 2.

(In the following Topic we will see that is equivalent to saying that polynomials are continuous functions. )

It is important to state again that when we write

Theorems on limits

the variable x is never equal to c, and therefore P(x) is never equal to P(c)exclamation Both c and P(c) are approached as limits.  The point is, we can name the limit simply by evaluating the function at c.

   Problem 4.   Evaluate   Theorems on limits

In that polynomial, let x = −1:

5(1) − 4(−1) + 3(1) − 2(−1) + 1
= 5 + 4 + 3 + 2 + 1
= 15.

   Problem 5.   Evaluate   Theorems on limits

On replacing x with c,  c + c = 2c.

   Problem 6.   Evaluate   Theorems on limits

[Hint:  This is a polynomial in t.]

On replacing t with −1:

3(−1)2 −5(−1) + 1 = 3 + 5 + 1 = 9.

   Problem 7.   Evaluate   Theorems on limits

[Hint:  This is a polynomial in h.]

On replacing h with 0, the limit is 4x3.

Some of the most important limits, however, will not be polynomials.

  They will be limits of certain quotients -- and they will appear to be  0
0
 Exclamation

Dealing with that will be the challenge.

Example 2.   Consider the function  g(x) = x + 2, whose graph is a simple straight line.  And just to be perverse (and to illustrate a logical point to which we shall return in Lesson 3), let the following function f(x) not be defined for x = 2.  That is, let

Theorems on limits

Theorems on limits

In other words, the point (2, 4) does not belong to the function; it is not on the graph.

Yet the limit as x approaches 2 -- whether from the left or from the right -- is 4Exclamation

Theorems on limits

For, every sequence of values of x that approaches 2, can come as close to 2 as we please.  (The limit of a variable is never a member of the sequence, in any case; Definition 2.1.)  Hence the corresponding values of f(x) will come closer and closer to 4.  Definition 2.2 will be satisfied.

End of the lesson

Next Lesson:  Continuous functions

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