Opening Problem
We will define xn as 0.333333… , with n 3s after the decimal point (e.g. so x2 = 0.33). n is restricted to positive integers.
| n | xn | 3xn | 1 – 3xn |
| 1 | |||
| 2 | |||
| 3 | |||
| 4 | |||
| 5 | |||
| 10 |
- Copy and complete the table above. In the last column, how many zeroes are there between the decimal point and the 1?
- Consider x100, which contains 100 3s. In the number 1 – 3x100, how many zeroes are there between the decimal point and the 1?
- As n gets progressively larger, how many zeroes are there in 1 – 3xn between the decimal point and the 1? How close does xn get to 1/3?
The opening problem shows us an example of a limiting process. As n tends to infinity the sequence approaches the limiting value 1/3. Another way of saying this is that for any number we might choose which is close to 1/3, we can find a value of n for which the sequence is closer and remains closer. We write .
We will use the following definition of a limit, which is sufficiently accurate for this course:
Definition of Limit: If f(x) is as close as we like to some real number A for all x sufficiently close to (but not equal to) a, then we say that f(x) has a limit of A as x approaches a, and we write .
The function may or may not be defined at x=a and this is irrelevant to the definition of the limit of f as x approaches a. We are interested in the behaviour of the function as x gets very close to a.
For instance, consider as
. If we substitute 0 in directly we get the meaningless value 0/0. However, if we rewrite the function as
we can see that for values close to zero f(x) = 5+x.
The graph of y=f(x) is the straight line y = x+5 with a hole, or a discontinuity, at (0,5). Even though f(0) doesn’t exist, the limit still exists, so as
. we write
.
Limit Laws
For functions f(x) and g(x) with and
, where a, l and m are real numbers, the following is true:
, provided
Worked Example
Evaluate the following:
Exercise


Answers
