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Limit and Continuity
Motivation: The limit of sequence and series has been introduced in the previous section. Based on that, we would like to turn the focus on functions first. A function maps one element from one set ( known as the domain of the function ) to the element of another set ( known s the range or image of the function ). Let’s start consider the following problem. Assume that a sequence converges to c . If we have a function , in other words, it maps a point from R (real number) to R, what will be the outcome if we ask the function f to map the whole sequence and observe the output of the function by taking the sequence as input ?
When we talk about , we should think about : for all the sequences converging to c, what is the behavior of the mapping of those sequences under the influence of the function f . However, for a point in one dimensional space, its surrounding points can only be classified in two directions.
Definition ( Limit from Below and Limit from Above ) The following notation represents that the limit of f from below b is d:
= d
That means while any sequence {xi , i N } approaches b from below, the sequence { f(xi), i N } approaches d. Similarly, the following notation represents that the limit of f from above b is c :
= c
If the two limits are equal, i.e. c=d , we can write
Definition ( Continuity ) A function f is continuous at the point b if
exits and .
In other words, for any > 0, there exists a corresponding >0 such that
| f(x) – f(b) | < when |x-b| <
What does this definition mean? We might get some idea from the following diagram: It means the curve f(x) is not “broken” at x=b . In the diagram above, f(x) has a “jump” at x=b. So, once you know how high the “jump” is and set less than that value, it is not going to satisfy what we define for “continuity” if you just choose the points around x=b from the other direction. Thus, f(x) in the diagram above is not continuous.
Example 1: f(x)=2x is continuous at x=1 .
Let’s consider in this way: Given | f(x)-f(1) | < t , where t > 0 -t < f(x) –f (1) < t -t < 2x – 2 < t -t/2 < x-1 < t/2 | x-1 | < t/2
So, for any >0, there exists a corresponding where = such that | f(x)-f(1)| < when | x-1 | <
Example 2: We define a function as below
That means: when x=1, f(x)=1; when x 1, f(x)=2x . Thus, , , but f(1)=2 .
f(x) is not continuous at x=1.
Example 3: In this example, f(x) is undefined when x=1 . However, and So, . But f(x) is not continuous at x=1 .
Theorem ( Intermediate Value Theorem ) Let f be a function such that and f is a continuous function. If f(a)f(b) < 0, then there must exist a point c between a and b such that f(c) = 0 .
Proof: Without loss of generality, let’s say a<b and f(a)<0, f(b)>0 . Let’s define two sequences { an } and { bn } inductively
a0=a, b0 = b, and ck = (ak + bk)/2 , k=0,1,2,3,4, …
for ak , bk where k=1,2,3,4,5, .., they are defined according to different conditions:
if f(ck-1) 0 , then ak = ak-1 and bk = ck ; if f(ck-1) < 0, then ak = ck-1 and bk = bk-1 . The reason of doing so is that we would like to find out the interval [ak, bk] such that f(ak) f(bk) < 0 and make the interval smaller and smaller. By doing that, we have a=a0 a1 a2 … an bn … b1 b0 = b and (bn – an ) = 2-n ( b-a ) for n=1, 2, 3, 4, 5, … f(an) 0 f(bn)
Can we say the sequences { an } and { bn } converge ? Recall the Cauchy criterion for sequence in R. Actually, this is known as the property of completeness of R . So, {an } and {bn} converge . Furthermore, = Hence, {an} and {bn} converge to the same limit. And f(x) is a continuous function, the implies f(an) and f(bn) also converge to the same limit. But f(an) 0 and f(bn) 0 , the limit of them must be 0 . In other words, there exists c in [an, bn] such that f(c) = 0 . QED.
Usually, this theorem has been introduced to find the root of a function, especially for polynomials . Later on, we will prove that all the polynomials are continuous functions. At this moment, we just use that result and check the following example.
Example: f(x)=x2 -2 . There exists c [1,2], such that f(c) = 0 .
Proof: f(1)= -1 < 0, f(2)=2 > 0 . Thus, from intermediate value theorem, we know that there exists c in [1,2] such that f(c)=0 .
Actually, we know it is . But how to represent the value of down to 6 digits after the decimal point ? This intermediate value theorem opens a door for that.
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