Chapter 10
Polynomial and Series Representations of Functions





10.1 Sums and Limits

10.1.2 More Notation for Limits

In Chapter 9 we introduced the notation

lim x f ( x )

for the limiting value of a function `f` as the independent variable goes to infinity. We noted in passing (in the Comment on Activity 1 for Section 9.2) that the same notation could be used for discrete sequences of numbers, such as those generated by Newton's Method for finding roots:

lim n t n = r ,

where `g` is a function, `r` is a root of `g text[(] t text[)] = 0`, and

t n + 1 = t n - g ( t n ) g ( t n ) .

In this case, the infinite sequence of numbers

t 0 , t 1 , t 2 , ...

is itself a function of the index variable `n`. That is, for each nonnegative integer `n` there is a unique value `t_n`. If Newton's Method converges, then the limiting value must in fact be a zero of `g`. We haven't added anything new here except notation for such limiting values.

Checkpoint 2Checkpoint 2

Example 2

Find the limiting value of the geometric sum

1 - 1 3 + 1 9 - 1 27 + ⋅⋅⋅ .

Solution   The ellipsis notation `cdots` indicates that this sum goes on forever. This is a sum of the form

1 + r + r 2 + r 3 + ⋅⋅⋅ ,

where `r = -1//3`. As we saw on the preceding page, the sum of the first `n` terms is

s n = 1 + r + r 2 + ⋅⋅⋅ + r n = 1 - r n + 1 1 - r .

Since `text[|] r text[|] <1`,

lim n r n + 1 = 0.

Thus the total sum is

end solution
lim n s n = lim n 1 - r n + 1 1 - r = 1 1 - r = 1 4 3 = 3 4 .

Limiting values have appeared in a number of other contexts, notably in the definition of the derivative. In this case, the independent variable `Delta t` approaches `0` rather than `oo`, but the notation works the same way:

f ( t ) = lim Δ t 0 f ( t + Δ t ) - f ( t ) Δ t .

This is a concise notation for a complicated process: At each fixed value of  `t`, and for each increment `Delta t` in `t`, form the difference quotient

change in values of f change in t ,

and take the limiting value as the change in `t` shrinks to `0`.

Example 3

Express the process of differentiating the exponential function in the notation for limiting values.

Solution   

end solution
d d t e t = lim Δ t 0 e t + Δ t - e t Δ t
  = e t lim Δ t 0 e Δ t - 1 Δ t .

Because we know the answer is `e^t` we also know the limiting value of the remaining quotient:

lim Δ t 0 e Δ t - 1 Δ t = 1.

Note that this last equation is actually an evaluation of `f' text[(] 0 text[)]`, where `f text[(] t text[)] = e^t`.

Checkpoint 3Checkpoint 3

There is nothing special about the variable `Delta t`. The limiting value notation works the same way no matter what the independent variable is — and no matter what number is being approached by that variable.

Activity 3

Find each of the following limiting values. Start by exploring each of the functions with your graphing tool. Once you have a value, try to explain why it must be the right value.

  1. lim x 2 x 3 - 8 x - 2

  2. lim x 0 ln ( 2 + x 2 ) x

  3. lim x 0 1 - cos x x

  4. lim x 0 1 - cos x x 2

Comment 3Comment on Activity 3

All of our examples of the use of the notation

lim x a f ( x )

have involved numbers `a` at which the function `f` is undefined. Functions have — or may have — limiting values at numbers in their domain as well. Usually it's not much of a challenge to find those limiting values: Just evaluate the function. That's because practically all our functions are continuous in the sense described in Chapter 7. Indeed, the definition we avoided giving there may be expressed in terms of limiting values: A function `f` is continuous at a number `a` if

lim x a f ( x ) = f ( a ) .

This simple-looking equation actually has three parts: First, it asserts that `a` is in the domain of `f`. That is, `f text[(] a text[)]` means something. Second, the values of `f text[(] x text[)]` approach something as `x` gets close to `a` — the same something on both sides, if `a` is not an endpoint of the domain. Third, the number being approached as limiting value and the number `f text[(] a text[)]` are the same. Finally, a function is continuous if it is continuous at each number in its domain. As we noted in Chapter 7, almost every formula-based function we see is continuous.

Example 4

Evaluate `lim_(x rarr 1) (x^3 - 8)/(x - 2)`.

Solution   The function `(x^3 - 8)/(x - 2)` is continuous at `1`. (In fact, it is continuous at every number except `2`.) Thus

end solution
lim x 1 x 3 - 8 x - 2 = 1 - 8 1 - 2 = 7.

Checkpoint 4Checkpoint 4

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