Prove that a function is continuous on an interval

In summary, the conversation discusses how to prove that the function \sqrt{x} is continuous on the interval [1,\infty). The conversation also explores the concept of uniform continuity and how to choose a suitable delta for proving continuity.
  • #1
DeadOriginal
274
2

Homework Statement


I have to prove that [itex]\sqrt{x}[/itex] is continuous on the interval [1,[itex]\infty[/itex]).2. The attempt at a solution
Throughout the school semester I believed that to show that a function is continuous everywhere all I need to do was show that [itex]\lim\limits_{h\rightarrow 0}f(x+h)-f(x)=0[/itex] and I never thought much about it. It never really came up as a problem on any of the homeworks or exams so I never had much problem with it. I am now doing review problems for the final and I realized that f(x)=1/x is a clear counterexample to what I stated above. The limit as h approaches 0 of f(x+h)-f(x) is 0 but 1/x is not continuous everywhere. Since this is the case, how do I show that a function is continuous everywhere within an interval?
 
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  • #2
To show it is continuous, you could go back to your first principles and use an epsilon delta argument.

Also given the interval, you can use the definition of the derivative, it's up to you.
 
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  • #3
Say I didn't know that [itex]\sqrt{x}[/itex] was differentiable. How would I use epsilon delta to prove that it was differentiable everywhere in that interval? The only epsilon delta proof for continuity that I know only proves that it is continuous at one point.

I read somewhere that I could choose my delta to not only depend on epsilon but also to depend on x. Is this correct? If this is, would it be right if I said

Choose [itex]\delta=\epsilon(|\sqrt{x}+\sqrt{a}|)[/itex]. Then [itex]|x-a|<\delta\Rightarrow|x-a|<\epsilon(|\sqrt{x}+\sqrt{a}|)\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}=|\sqrt{x}-\sqrt{a}|<\epsilon[/itex].
 
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  • #4
DeadOriginal said:
Say I didn't know that [itex]\sqrt{x}[/itex] was differentiable. How would I use epsilon delta to prove that it was differentiable everywhere in that interval? The only epsilon delta proof for continuity that I know only proves that it is continuous at one point.

I read somewhere that I could choose my delta to not only depend on epsilon but also to depend on x. Is this correct? If this is, would it be right if I said

Choose [itex]\delta=\epsilon(|\sqrt{x}+\sqrt{a}|)[/itex]. Then [itex]|x-a|<\delta\Rightarrow|x-a|<\epsilon(|\sqrt{x}+\sqrt{a}|)\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}=|\sqrt{x}-\sqrt{a}|<\epsilon[/itex].

No, that isn't right. When you are discussing continuity at the point ##x=a##, the ##\delta## will normally depend on ##\epsilon## and ##a##. If it doesn't depend on ##a## you have uniform continuity. So in your example, ##\delta## can depend on the ##a## but not ##x##. But you have$$
|\sqrt x -\sqrt a| = \frac {|x-a|}{|\sqrt x + \sqrt a|}$$Given that you have ##x\ge 1## can you find an overestimate of the right side that doesn't involve ##x## in the denominator, then choose ##\delta## to make it work?
 
  • #5
LCKurtz said:
No, that isn't right. When you are discussing continuity at the point ##x=a##, the ##\delta## will normally depend on ##\epsilon## and ##a##. If it doesn't depend on ##a## you have uniform continuity. So in your example, ##\delta## can depend on the ##a## but not ##x##. But you have$$
|\sqrt x -\sqrt a| = \frac {|x-a|}{|\sqrt x + \sqrt a|}$$Given that you have ##x\ge 1## can you find an overestimate of the right side that doesn't involve ##x## in the denominator, then choose ##\delta## to make it work?

I'm looking at ##|x-a|<\delta## and I'm thinking that the way I would find a bound for for ##|\sqrt{x}|## is to first make ##\delta_{1}=1## so that I have ##|x-a|<1\Rightarrow -1<x-a<1\Rightarrow a-1<x\Rightarrow \sqrt{a-1}+\sqrt{a}<\sqrt{x}+\sqrt{a}\Rightarrow \frac{1}{\sqrt{x}+\sqrt{a}}<\frac{1}{\sqrt{a-1}+\sqrt{a}}## and since these numbers are positive ##\frac{1}{|\sqrt{x}+\sqrt{a}|}<\frac{1}{|\sqrt{a-1}+\sqrt{a}|}##
What if I then chose ##\delta=\min(\delta_{1},\epsilon(|\sqrt{a-1}+\sqrt{a}|))##?

Then my epsilon delta argument would be:
Choose ##\delta=\min(\delta_{1},\epsilon(|\sqrt{a-1}+\sqrt{a}|))##. Then
##|x-a|<\delta\Rightarrow |x-a|<\epsilon(|\sqrt{a-1}+\sqrt{a}|)\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon##.

Also while we are talking about uniform continuity, if delta doesn't depend on ##a## for uniform continuity then what does delta depend on? Does it depend on x and y in that case? Like say I wanted to prove that ##\sqrt{x}## was uniformly continuous on [1,##\infty##) and I have ##|x-y|<\delta##, can I then choose a ##\delta## that depends on ##x## and ##y## to show that ##|\sqrt{x}-\sqrt{y}|<\epsilon##?
 
  • #6
DeadOriginal said:
I'm looking at ##|x-a|<\delta## and I'm thinking that the way I would find a bound for for ##|\sqrt{x}|## is to first make ##\delta_{1}=1## so that I have ##|x-a|<1\Rightarrow -1<x-a<1\Rightarrow a-1<x\Rightarrow \sqrt{a-1}+\sqrt{a}<\sqrt{x}+\sqrt{a}\Rightarrow \frac{1}{\sqrt{x}+\sqrt{a}}<\frac{1}{\sqrt{a-1}+\sqrt{a}}## and since these numbers are positive ##\frac{1}{|\sqrt{x}+\sqrt{a}|}<\frac{1}{|\sqrt{a-1}+\sqrt{a}|}##
What if I then chose ##\delta=\min(\delta_{1},\epsilon(|\sqrt{a-1}+\sqrt{a}|))##?

That just makes my head hurt. Why don't you just use my suggestion, since ##x\ge 1##:$$
\frac {|x-a|} {\sqrt x + \sqrt a }\le \frac {|x-a|} {1+\sqrt a}$$and use that to figure out ##\delta##?
Then my epsilon delta argument would be:
Choose ##\delta=\min(\delta_{1},\epsilon(|\sqrt{a-1}+\sqrt{a}|))##. Then
##|x-a|<\delta\Rightarrow |x-a|<\epsilon(|\sqrt{a-1}+\sqrt{a}|)\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon##.

Also while we are talking about uniform continuity, if delta doesn't depend on ##a## for uniform continuity then what does delta depend on? Does it depend on x and y in that case?
No. In that case ##\delta## depends only on ##\epsilon##, not ##x## or ##y##. Look at my suggestion above and see if you can also overestimate getting rid of the ##a## in the denominator to get a ##\delta## that depends only on ##\epsilon##.
Like say I wanted to prove that ##\sqrt{x}## was uniformly continuous on [1,##\infty##) and I have ##|x-y|<\delta##, can I then choose a ##\delta## that depends on ##x## and ##y## to show that ##|\sqrt{x}-\sqrt{y}|<\epsilon##?

No.
 
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  • #7
LCKurtz said:
That just makes my head hurt. Why don't you just use my suggestion, since ##x\ge 1##:$$
\frac {|x-a|} {\sqrt x + \sqrt a }\le \frac {|x-a|} {1+\sqrt a}$$and use that to figure out ##\delta##?

OOOOOOOOOOOhhhhhhhh. So since ##\frac {|x-a|} {\sqrt x + \sqrt a }\le \frac {|x-a|} {1+\sqrt a}## we can choose ##\delta=\epsilon|1+\sqrt{a}|## and then get

##|x-a|<\epsilon|1+\sqrt{a}|\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}<\epsilon|1+\sqrt{a}|\frac{1}{|1+\sqrt{a}|}\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon##. Third times a charm?

No. In that case ##\delta## depends only on ##\epsilon##, not ##x## or ##y##. Look at my suggestion above and see if you can also overestimate getting rid of the ##a## to get a ##\delta## that depends only on ##\epsilon##.

Hmm. Thanks. This helps to clarify a lot! I will do this part on my own time. As long as I get the top part, it seems like this part shouldn't be too much harder to do.
 
  • #8
DeadOriginal said:
OOOOOOOOOOOhhhhhhhh. So since ##\frac {|x-a|} {\sqrt x + \sqrt a }\le \frac {|x-a|} {1+\sqrt a}## we can choose ##\delta=\epsilon|1+\sqrt{a}|## and then get

##|x-a|<\epsilon|1+\sqrt{a}|\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}<\epsilon|1+\sqrt{a}|\frac{1}{|1+\sqrt{a}|}\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon##. Third times a charm?



Hmm. Thanks. This helps to clarify a lot! I will do this part on my own time. As long as I get the top part, it seems like this part shouldn't be too much harder to do.

Good, now you are getting somewhere. Your ##\delta## depends on ##\epsilon## and ##a##, so you have continuity for each ##a##. Now, can you find a ##\delta## that doesn't depend on ##a## either? If it depended only on ##\epsilon## you would have shown uniform continuity.
 
  • #9
LCKurtz said:
Good, now you are getting somewhere. Your ##\delta## depends on ##\epsilon## and ##a##, so you have continuity for each ##a##. Now, can you find a ##\delta## that doesn't depend on ##a## either? If it depended only on ##\epsilon## you would have shown uniform continuity.

As per your advice I noted that the smallest value of ##a## would also be 1 on the interval so choosing ##\delta=2\epsilon## I would have:

##|x-a|<\delta\Rightarrow |x-a|<2\epsilon\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}<2\epsilon\frac{1}{2}\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon##.

Thank you! This has clarified a lot for me!
 
  • #10
DeadOriginal said:
As per your advice I noted that the smallest value of ##a## would also be 1 on the interval so choosing ##\delta=2\epsilon## I would have:

##|x-a|<\delta\Rightarrow |x-a|<2\epsilon\Rightarrow |x-a|\frac{1}{|\sqrt{x}+\sqrt{a}|}<2\epsilon\frac{1}{2}\Rightarrow |\sqrt{x}-\sqrt{a}|<\epsilon##.

Thank you! This has clarified a lot for me!

OK, you have all the pieces figured out, but you could present it a little more smoothly. Start with
$$
|\sqrt x - \sqrt a| = \frac{|x-a|}{\sqrt x + \sqrt a}\le \frac{|x-a|}{2}$$since ##x\ge 1## and ##a\ge 1##. (This is really the "exploratory argument" where you are figuring out what ##\delta## might work.)

Suppose ##\epsilon>0##. Let ##\delta=2\epsilon##. Then if ##|x-a|<\delta = 2\epsilon## we have$$
|\sqrt x - \sqrt a| <\frac{2\epsilon} 2=\epsilon$$
 

1. What does it mean for a function to be continuous on an interval?

Continuity of a function on an interval means that the function is defined and has a value at every point within the interval, and that the function's value changes smoothly and without abrupt jumps or breaks.

2. How can I prove that a function is continuous on an interval?

To prove continuity on an interval, you need to show that the limit of the function as x approaches any point within the interval is equal to the function's value at that point. This can be done using the epsilon-delta definition of continuity or by using the intermediate value theorem.

3. What is the epsilon-delta definition of continuity?

The epsilon-delta definition of continuity states that a function f(x) is continuous at a point x=a if for any positive number ε, there exists a positive number δ such that for all x within δ units of a, the output of the function f(x) is within ε units of f(a).

4. Can a function be continuous on an open interval but not on a closed interval?

Yes, a function can be continuous on an open interval but not on a closed interval. This is because continuity at the endpoints of a closed interval is not guaranteed, while continuity at the endpoints of an open interval is not required.

5. Are there any common types of functions that are always continuous on an interval?

Yes, there are several common types of functions that are always continuous on an interval. These include polynomial functions, rational functions, exponential functions, logarithmic functions, and trigonometric functions.

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