2

LIMITS

A sequence of rational numbers

The limit of a variable

Left-hand and right-hand limits

The definition of the limit of a variable

The limit of a function

Theorems on limits

Limits of polynomials

CENTRAL TO CALCULUS is the idea of a limit.  And central to the idea of a limit is the idea of a sequence of rational numbers.

Every irrational number is in fact the limit of a sequence of rational numbers. And again, the limit, the irrational number, is not a member of the sequence.

We speak of a sequence as being infinite, which is a brief way of saying that, no matter how many terms we have written, we could always write one more. And when the sequence has a limit, each term we add will be closer to the limit than the previous term.

When a variable x approaches a limit l, we symbolize that as  x l. Read:  "The values of x approach l as a limit," or simply, "x approaches l." In the example above,  x 2.  "x approaches 2."

We also say that a sequence converges to a limit.  The above sequence converges to 2.

In what follows, by a sequence we mean a sequence of rational numbers.

Now the sequence we chose were values less than 2.  Hence we say that x approaches 2 from the left.  We write

x 2−

But we can easily construct a sequence that converges to 2 from the right; that is, a sequence of values that are more than 2.  For example,

2.2,  2.1,  2.01,  2.001,  2.0001,  2.00001, .  .  .

In this case, we write  x 2+ .

But again, no matter what small number we name, if we take enough terms in that sequence, the values of the difference |x − 2| will become and remain less than that small number.

We have defined the limit of a variable, but what we typically have is a function of a variable -- which is also a variable.  For example,

y = f(x) = x².

Now, a sequence of values of x will force a sequence of values of f(x). The question is:  If the values of x approaches a limit, will the corresponding values of f(x) also approach a limit?  If that is the case -- if f(x) approaches a limit L when x approaches a limit l -- then we write

"The limit of f(x) as x approaches l, is L."

In fact, let us see what happens to  f(x) = x²  as x 2−.  Suppose again that x assumes this sequence of values:

1.9,  1.99,  1.999,  1.9999,  1.99999, . . .

x² will then assume this sequence:

(1.9)²,  (1.99)²),  (1.999)²,  (1.9999)²,  (1.99999)², . . .

It is easy to see that x² approaches 2² = 4.

Again, this means that if took enough terms in the sequence of differences, x² − 4 --

(1.9)² − 4,   (1.99)² − 4,   (1.999)² − 4,   (1.9999)² − 4, . . .

-- then the absolute values of any terms we might then write, would be less than any positive number we name, however small.

Moreover, if we consider a sequence  x 2+:

2.2,  2.1,  2.01,  2.001,  2.0001,  2.00001, .  .  .

then terms of the sequence  x² − 4  --

(2.2)² − 4,   (2.1)² − 4,   (2.01)² − 4,   (2.001)² − 4,  .  .  .

-- again will eventually be less than any small number.

4 is the limit of f(x) whether x approaches 2 from the left or from the right.  Therefore we can drop the + or − and simply write:

To summarize:

In other words, for the limit of f(x) to exist as x approaches l , the left-hand and right-hand limits must be equal.

if and only if

We will see that this definition of the limit of a function existing as x approaches l, becomes the definition of the function being continuous at that value -- if L is the value of the function there; that is, if L = f(). (Definition 3.)

The most important limit -- the limit on which calculus hinges -- is called the derivative. All the other limits studied in Calculus I are logical fun and games, never to be heard from again.

For example,

 

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

1)			(f + g)  =  A + B
2)			(f g)  =  AB

3)

f g

=

A B

,  if B is not 0.

Also, if c is a constant -- that is, independent of x -- then

4)

 

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 a constant

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

5)

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

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

Solution.   x² = x· x.  And we have proved that

exists, and is

equal to 4.  Therefore,

according to Theorem 2.

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

Problem 2.

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

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

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

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

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

The student might think that to evaluate a limit all we do is evaluate the function at that value.  And for the most part that is true  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:

P(x) = a_nxⁿ + a_n−1xⁿ⁻¹ + . . . + a₁x +a₀

where n is a whole number, and a_n0.

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

(We will see that this is equivalent to saying that polynomials are continuous functions.  Definition 3.)

x is never equal to c, and P(x) is never equal to P(c)  Both c and P(c) are approached as limits.  "Closer and closer."  Nevertheless, we can evaluate the limit simply by evaluating the function at c.

Problem 4.   Evaluate

In that polynomial, let x = −1. Then

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

Problem 5.   Evaluate

On replacing x with c,  c + c = 2c.

Problem 6.   Evaluate

[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

[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

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

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 4

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.

Next Lesson:  Continuous functions

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