Real Analysis – Limits and Continuity VI
— More properties of continuous functions —
Definition 35
Let ; and . If , we can define as: |
As an application of the previous definition let us look into . It is .
Since we can define as
As another example let us look into Since and we can’t define for .
Finally if we let we have . Since the limits are divergent we still can’t define .
In general one can say that given and exists if and only if exists and is finite.
Theorem 42 Let ; and . If and are continuous functions then , and (if ) are also continuous functions.
Proof: We’ll prove that is continuous and let the other cases for the reader. Let be a sequence of points in such that . Then and (since and are continuous functions). Hence it follows from property of Theorem 19. Which is the definition of a continuous function. |
Let . First we note that , and are continuous functions. Now also a continuous function. is continuous since it is the product of continuous functions. is continuous since it is the product of continuous functions. Finally is continuous since it is the sum of continuous functions.
As an application of the previous theorem let . Since we can write . Since is a continuous function and is also a continuous function it follows that is a continuous function (it is the composition of two continuous functions).
By the same argument we can also show that with , (for ) is also a continuous function in .
Theorem 44 Let , , and . Suppose that and that exists. If is continuous it follows .
Proof: Omitted. |
Find .
We can write . Since it is, from Theorem 44, .
In general if it is . In conclusion
Suppose that and let be the function that makes be continuous in .
It is , hence it is .
By definition is continuous so by Theorem 44 .
Thus we can conclude that when it is
For example .
Let . By Theorem 44 it is (with the conventions and ). Thus .
Analogously one can show that (with the conventions and ).
Let . It is (since ).
On the other hand for it also is .
The question we want to answer is since the answer to this question tell us which of the functions tends more rapidly to .
Theorem 45 Let and . Then
Proof: Let (). It is . Hence . Moreover . which is Let denote the nearest integer function and using Bernoulli’s Inequality () it is . Hence . Since it follows from Theorem 32 that . Using 18 and setting it is |
Corollary 46 Let , then
Proof: Left as an exercise for the reader (remember to make the convenient change of variable). |
Theorem 47 Let , then =0. Proof: First remember that which is Stirling’s Approximation.
Since it also is . And Thus For the argument of the exponential function it is
Hence . |
Lemma 48
Proof: Omitted. |
Theorem 49 |
Corollary 50
Proof: Left as an exercise for the reader. Make the change of variables and use the previous theorem. |
Generalizing the previous results one can write with full generality:
- if
- if
- if
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