What are the requirements for a function to be continuous at a point?

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rbzima
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Hey guys, I'm in a little bit of a jam here:
I managed to miss a really important lecture on continuity the other day, and there were a few examples that the professor provided to the class that I just got, but would love it if someone could explain them to me.

First, f(x)=x3 is continuous on all of R. Everyone went through this, but I'm not really sure how it works.

Second, we used the fact that "a function [tex]f:A \rightarrow[/tex] R fails to be uniformly continuous on A if and only if there exists a particular [tex]\epsilon_{0} > 0[/tex] and two sequences (xn) and (yn) in A satisfying

[tex]\left| x_n - y_n\right| \rightarrow 0[/tex] but [tex]\left| f(x_n) - f(y_n)\right| \rightarrow \epsilon_0[/tex]

to show fis not uniformly continuous on R.

Thirdly, we showed that f is uniformly continuous on any bounded subset of R
 
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I studied continuity in the last semester. I can't remember all the theorems ATM.
Any way, as far as I can remember that a function, in plain words, can be continuous
1) at a point or,
2) can be continuous on a certain interval.

For continuity at a point it must have a value at that point
For an interval, it must be continuous at every points in the interval.

I can remember upto this.
 
rbzima said:
Hey guys, I'm in a little bit of a jam here:
I managed to miss a really important lecture on continuity the other day, and there were a few examples that the professor provided to the class that I just got, but would love it if someone could explain them to me.

First, f(x)=x3 is continuous on all of R. Everyone went through this, but I'm not really sure how it works.

Second, we used the fact that "a function [tex]f:A \rightarrow[/tex] R fails to be uniformly continuous on A if and only if there exists a particular [tex]\epsilon_{0} > 0[/tex] and two sequences (xn) and (yn) in A satisfying

[tex]\left| x_n - y_n\right| \rightarrow 0[/tex] but [tex]\left| f(x_n) - f(y_n)\right| \rightarrow \epsilon_0[/tex]

to show fis not uniformly continuous on R.

Thirdly, we showed that f is uniformly continuous on any bounded subset of R

For the first quesiton, you can use the contunuity of limit. As I remember ıt is something like that. lim(x --> a) f(x) = f(a). and for f(x^3), this statement is correct.
 
mdnazmulh said:
I studied continuity in the last semester. I can't remember all the theorems ATM.
Any way, as far as I can remember that a function, in plain words, can be continuous
1) at a point or,
2) can be continuous on a certain interval.

For continuity at a point it must have a value at that point
For an interval, it must be continuous at every points in the interval.

I can remember upto this.

This is incorrect.

One counter example is the following function

Let [tex]f(x)=\left\{\begin{array}{cc}0,&\mbox{ if } <br /> x\neq 0\\1, & \mbox{ if } x=0\end{array}\right.[/tex]

This function has a value at x = 0, but is certainly not continuous at that point.

From Wiki,
To be more precise, we say that the function f is continuous at some point c when the following two requirements are satisfied:

f(c) must be defined (i.e. c must be an element of the domain of f).

The limit of f(x) as x approaches c must exist and be equal to f(c).
 
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