Proving Function Continuity in [-1,30]: Understanding the Example

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In summary, the conversation discusses the process of proving continuity for a function in a given interval. The individual is struggling to understand the connection between the interval and the solution, but eventually realizes that the function must be continuous for all values in its prescribed domain. The conversation also touches on the use of limits and the definition of continuity.
  • #1
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I am trying to fully understand this example from a textbook I am reading:

http://img59.imageshack.us/img59/9237/continuityyn8.jpg

What I am not understanding is how they are proving it for [-1,1].. The way I see it is they proved that the function is continuous for all values in it's domain...

For example, I thought up this problem on my own to help me understand :

Given [tex] f(x)=1-\frac{1}{x-4} [/tex], prove that f(x) is continuous in the interval [-1,30] (Obviously it's not continuous at x=4.) The problem is that I don't see the connection between the interval and the solution...

I can just go ahead and prove that [tex]\lim_{x\rightarrow a}f(x)=f(a)[/tex]... Which was stated in my text as meaning that the function is continuous... which it obviously isnt.

[tex]\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}(1-\frac{1}{x-4})[/tex]
[tex]=1-\lim_{x\rightarrow a}\frac{1}{x-4}[/tex]
[tex]=1-\frac{1}{a-4}[/tex]
[tex]=f(a)[/tex]

Can someone cure my confusion? Thanks guys.
 
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  • #2
I am not sure as to whether or not the latex is really screwed up in my post, I edit it and it looks completely diff from what I see when I refresh the post. Click on the indivual latex box's to see what I meant to type if it comes up screwy. Maybe it's just my pc.

Nevermind, it's fine now.
 
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  • #3
oh my.. I see it now, limf(x)=f(a) for all values except a=4... since that would result in an undefined statement. But I still don't understand how they are testing for the interval [-1,1]... How would I test my made up example in the interval [-1,1]?
 
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  • #4
They are, indeed, showing it for the values in the function's PRESCRIBED DOMAIN, that is the interval from -1 to 1
 
  • #5
The only time they implicity use the assmption that x was in [-1,1] was in moving the limit inside the square root. sqrt(x) is only continuous for x>=0, (namely because it is only defined here), so they needed that fact that 1-x^2 was non-negative.
 
  • #6
Okay I understand what they were doing now, thanks :)

edit:Nevermind this other question, I reread the definition/example and it became clear.
 
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Related to Proving Function Continuity in [-1,30]: Understanding the Example

1. What is function continuity?

Function continuity is the property of a function where small changes in the input variable result in small changes in the output variable. This means that the function is smooth and has no sudden jumps or gaps in its graph.

2. How is function continuity proven in the example?

In the example, function continuity is proven by showing that the function is defined and continuous at every point within the given interval of [-1,30]. This is done by evaluating the limit of the function at each point and ensuring that it is equal to the value of the function at that point.

3. What is the significance of the interval [-1,30] in proving function continuity?

The interval [-1,30] is used to prove function continuity because it includes all the points where the function may have discontinuities. By evaluating the limit of the function at every point within this interval, we can determine if the function is continuous at all points within the interval.

4. What is the importance of proving function continuity?

Proving function continuity is important in mathematics and science because it helps us understand and analyze the behavior of functions. It allows us to make accurate predictions and use mathematical models to solve real-world problems.

5. What are some common methods for proving function continuity?

Some common methods for proving function continuity include using the definition of continuity, evaluating one-sided and two-sided limits, and using the intermediate value theorem. Other methods such as the squeeze theorem and the delta-epsilon definition can also be used in certain cases.

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