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.

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.  Thus the above sequence converges to 2.

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 go far enough out in the sequence |x − 2|, the values of that difference ultimately will become less than that small number.

Again, when we say that a variable x approaches a "limit" l, we mean that the values of x get closer and closer to l.  Yet the values of x are never equal to l.

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 we go far enough out 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 all subsequent terms eventually will 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 from the left or from the right.  Therefore we can drop the + or − and simply write:

To summarize:

For every sequence of values of x that approach as a limit -- whether from the left or from the right -- the corresponding values of f(x) approach L as a limit. (Definition 2.1.)

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

We will see that this definition of the limit existing at a point, becomes the definition of the function being continuous at that point -- if L is the value of the function at that point; that is, if L = f(). (Definition 3.1.)

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.

Now here is an example of a function not approaching a limit:

As x approaches 0 -- as it becomes a very small number -- its

larger than any number we name.  It does not approach a limit.

Let us 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.)

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.

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

4)

 

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 but f depends on x, then

5)

A constant 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, we 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 and Theorem 4.

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

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 point.  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 with argument x has this general form:

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 a polynomial as the argument 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 3.1.)

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 that point.

Problem 4.   Evaluate

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.

Making sense of that will be our challenge.

Example 2.   Consider the function  f(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 f(x) not exist when 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 satsisfied.

Next Lesson:  Continuous functions

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