— 6. Limits and Continuity —
After introducing sequences and gaining some knowledge of some of their properties (I, II, III, and IV) we are ready to embark on the study of real analysis.
— 6.1. Preliminary Definitions —
Physics is expressed best and most powerfully in the language of mathematics and a very useful mathematical concept for physics is the concept of a function.
Generally speaking a function is an association (it transforms an input signal from the first set into an output signal of a second set) between the elements of two sets.
The sequences we studied are a special case of functions: they take natural numbers (or a subset of them) as their input signals and map them to real numbers.
Now, more formally we introduce:
Sometimes we may not be interested in how the function maps the whole of $ {D}$ but just on a particular subset of $ {D}$. So it makes sense to introduce:
| Definition 24 Given $ {E \subset D}$ it is $ {f\left[ E \right] = \left\lbrace f(x):x \in E \right\rbrace }$ is the image of $ {f}$ by $ {E}$. |
As we did for sequences we can too define what is a bounded above function, a bounded below function, a bounded function and etc.
As an example we'll give:
| Definition 25 $ {f}$ is said to be bounded iff $ {\exists \, \alpha > 0 : |f(x)| \leq \alpha \forall x \in D }$ |
— 6.2. Introduction to Topology —
Once again so that we don't let things get too abstract let us give an example:
$ \displaystyle E = \left] 0,1\right[ \cup \left\lbrace 2 \right\rbrace $
It is easy to see (and we won't give a rigorous proof of that) that $ {E^\prime= \left[ 0,1 \right] }$ and that $ {2}$ is the only isolated point of $ {E}$.
| Definition 28 The symbol $ {D_{c^+}}$ will be used to denote $ {D \cap \left] c, \infty \right[ }$ and the symbol $ {D_{c^-}}$ will denote $ {D \cap \left] - \infty , c \right[ }$ |
As an example let us calculate
$ \displaystyle \lim _{x \rightarrow 0^+} \frac{1}{x} $
In this case it is $ {D_{0^+} = \left] 0, \infty \right[ }$ and $ {0^+ \in D^\prime_{c^+}}$ so that the limit we intend to calculate indeed makes sense.
If $ {x_n}$ is a sequence of points in $ {D^\prime_{c^+}}$ such as $ {x_n \rightarrow 0^+}$ then it follows that $ {\lim f(x_n)=\lim \dfrac{1}{x_n}=\dfrac{1}{0^+}=+\infty }$
As an application of theorem 28 let us calculate the following limit
$ \displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x} $
It is easy to see that this limit doesn't exist. Let $ {f(x)=\dfrac{1}{x}}$ it is $ {\displaystyle \lim_{x \rightarrow 0^+} f(x) = +\infty}$ and $ {\displaystyle \lim_{x \rightarrow 0^-} f(x) = -\infty}$.
Since the limit from the left is different from the limit from the right we can conclude that $ {\displaystyle\lim_{x \rightarrow 0}\dfrac{1}{x}}$ doesn't exist.
| Definition 29 $ { +\infty }$ is a limit point of $ {E}$ if $ {E}$ isn't bounded above in $ { \mathbb{R} }$. $ { -\infty }$ is a limit point of $ {E}$ if $ {E}$ isn't bounded below in $ {\mathbb{R}}$. |
If you're having trouble understanding these definitions just think that if $ {E}$ isn't bounded above than it means that $ { \exists x_n \in E: \quad \lim x_n = +\infty }$.
Which is just the definition of limit point.
| Definition 30 $ {c}$ is said to be a limit point of $ {E}$ if $ \displaystyle \forall \delta > 0 \; V(c,\delta) \cap E \setminus \left\lbrace c \right\rbrace \neq \emptyset$ |
We'll only define the limit of a function in limit points of the domain. Notice that by this way we can too define the limit of points that don't belong in the domain of the function.
As always a few examples will be provided in order for us to test our knowledge.
- Calculate the limit of $ {\displaystyle \lim_{x \rightarrow + \infty} \dfrac{1}{x} }$.
$ { D = \mathbb{R} \setminus \left\lbrace 0 \right\rbrace }$ and $ { + \infty \in D^\prime }$ since $ {D}$ isn't bounded above in $ { \mathbb{R} }$. Thus the limit we set ourselves to calculate makes sense in our theory of limits.
Let $ {x_n}$ be a sequence of points in $ {D}$ such as $ { x_n \rightarrow + \infty }$ and $ {f(x)=\dfrac{1}{x}}$, then $ {f(x_n)=\dfrac{1}{x_n}}$ and it always is $ {\lim f(x_n)=0}$.
- Calculate the limit of $ {\displaystyle \lim_{x \rightarrow + \infty} \sin x }$
Choosing $ {f(x)= \sin x}$ we see that the domain is $ {D = \mathbb{R}}$. Thus $ {+\infty \in D^\prime}$
Let us choose $ {x_n = n \pi}$. Thus $ {x_n \rightarrow +\infty }$ and $ {f(x_n)=\sin x_n = 0}$.
In this case it trivially is $ {\lim f(x_n)=0}$.
Now if we choose $ {y_n=\pi/2 + 2n\pi}$ it also is $ {y_n \rightarrow + \infty}$ but $ {f(y_n)= \sin (\pi/2+2n\pi)=1}$ and so $ {\lim f(y_n)=1}$.
Thus we were able to find $ {x_n}$, $ {y_n}$ such as $ {\lim x_n = \lim y_n = + \infty}$ but $ {\lim f(x_n) \neq \lim f(y_n)}$. Thus $ {\displaystyle \lim_{x \rightarrow +\infty} \sin x }$ doesn't exist.
Formalizing the previous notions:
Post a Comment