# Proof f(x) = x is continuous

1. Jun 4, 2010

### Moogie

Hi

I'm new to calculus and I'm teaching myself so please be kind to me :)

How do you prove that f(x) = x is continuous at all points? I know a little bit about deltas and epsilons. I know that for a positive epsilon it is possible to get a delta such that

|(fx) - L| < epsilon for all x that satisfy 0 < |x-a| < delta

but i don't know if that can be applied here

I know that for f(x) = x then if it is continuous then you have to prove

lim (x->a) (fx) = f(a) for all x (i know the definition is more precise)

You can rewrite
lim (x->a) (fx) = f(a)

as

lim (x->a) x= a

but this makes sense intuitively but is it a proof? How could you show this with deltas and epsilons.

I hope this isn't asking too much as I don't think its a hard answer for someone who knows what they are talking about

thanks

2. Jun 4, 2010

### Office_Shredder

Staff Emeritus
This is one of those problems where the answer comes so fast from the definition that it's hard to see that you actually found it

The basic definition: For some value a, given $$\epsilon$$, we need to prove there exists $$\delta$$ such that if $$|x-a|<\delta$$, then $$|f(x)-f(a)|<\epsilon$$. The first step is to substitute f(x) into this: f(x) is just x, and f(a) is just a.

For some value a, given $$\epsilon$$, we need to prove there exists $$\delta$$ such that if $$|x-a|<\delta$$, then $$|x-a|<\epsilon$$

So if I tell you that $$|x-a|<\epsilon$$ is required, what value of $$\delta$$ should you pick?

3. Jun 4, 2010

### Moogie

I think this is where i got stuck when i tried to get the answer myself...

Why did you write |f(x) - f(a)| instead of |f(x) - L|

I get the feeling i'm not seeing the wood for the trees here. Is it because if
f(x) is continuous, then f(x) goes to f(a) as x goes to a, so the limit at a is f(a)?

But how can you say the limit at a is f(a)? Isn't that what you are trying to prove?

Or are you going to assert it and then prove it by saying you can get a delta to satisfy the epsilon for it, if that makes sense

4. Jun 4, 2010

### Tedjn

Good question. You are trying to prove that the limit of f(x) as x goes to a is f(a), because that is the condition for continuity. The actual definition says that L is a limit if ... epsilons and deltas .... If you substitute f(a) (which exists independent of what the limit is) for L, you get that f(a) is a limit if ... epsilons and deltas ..., which is what you want to show.

5. Jun 4, 2010

### Moogie

Is that how you prove a limit then? Or at least one way to prove a limit.

You assert what the limit is and if you can show that for every positive epsilon there is a delta such that |f(x)-L| < epsilon for |x -a| < delta?

That seems backwards to me, or perhaps that is the way mathematical proofs are done i.e. you state the answer first and show it is true rather than 'deriving' the answer at the end from a series of steos.

6. Jun 4, 2010

### Staff: Mentor

Yes, that's right.
It's not really backward if you think about it. When you prove that the limit of a function is L, you have to already know what the limit is. You can think of this as being two parts: determining the limit, and then proving beyond doubt that the value you found is correct.

7. Jun 4, 2010

### Moogie

@officeshredder (or anyone)

so we assert that the limit is f(a). For this to be true we have to show that for every positive epsilon there is a delta such that |f(x)-L| < epsilon for |x -a| < delta?

so |f(x) - f(a)| < epsilon for all |x-a| < delta
or |x-a| < epsilon for |x-a| < delta

How do i start picking deltas and epsilons? This is my thought process:

Lets say i pick an epsilon greater than 0. This means the output of the function lies in the range (a-epsilon, a + epsilon). This means i have to pick a delta such that if |x-a| < delta, the outputs lie in that range.

I have no idea how to show this or test this

8. Jun 4, 2010

### Tedjn

You are looking for a way to choose delta based on what epsilon is. For example, suppose that δ = g(ε). Then, if your choice of g(ε) is appropriate, you would have |x-a| < ε whenever |x-a| < g(ε).

9. Jun 4, 2010

### Moogie

If delta is less than epsilon then every input in the range (a-delta, a+delta) gives and output less than (a-epsilon, a+epsilon)?

10. Jun 4, 2010

### Tedjn

Yes, that's right. Common practice is to do the least amount of work as possible, so people will often just choose one delta from the range of possible deltas you mentioned, e.g. δ = ε/2. But this problem is simple enough that your answer is perfectly all right, even better.

11. Jun 4, 2010

### Moogie

It's delta less than or equal to epsilon?

My brain keeps confusing me and saying to me 'how do i know |x-a| < epsilon and then i get muddled.

Is this a common source of confusion? Presumbaly the answer is because i set epsilon such that |x-a| < epsilon but that's where i keep wandering off track and losing focus

12. Jun 4, 2010

### Staff: Mentor

For this function, you can take delta to be equal to epsilon.

A delta epsilon proof can be thought of as a dialog between someone who doubts that L is the limit and someone who intends to prove that L is the limit.

The doubter says, "Well can you get f(x) within .05 of L?"
You say, "Sure, if x is within delta of a, then f(x) will be within .05 of L.

The doubter then says, "But can you get f(x) within .001 of L?"
You say, "Sure, I'll just choose a smaller delta, then f(x) will be within .001 of L.

and so on. Eventually, the doubter realizes that no matter how small an epsilon he tells you, you are able to find a number delta so that for any x within delta of a, f(x) will be within epsilon of L. On making this realization, the doubter gives up, and cedes the argument to you.

The real power of this technique is that it can be used even when the function isn't defined at a; i.e., f(a) doesn't exist.

13. Jun 4, 2010

### Tedjn

Oh, I confused myself there. You can set delta less than or equal to epsilon (so actually δ = ε works just fine even though δ = ε/2 works as well).

This is a common point of confusion. Think about what it means to set δ = ε. We want to ascertain that for every ε > 0 there exists δ such that ... So, is this true if we set δ = ε? Yes, since setting δ = ε we get whenever |x - a| < δ = ε, |x - a| < ε.

14. Jun 4, 2010

### Moogie

This made a lot of sense to me - thank you

But the bit that kept confusing me when I was trying to work this out was

"how do i know |x-a| < epsilon. I don't even know what epsilon is so how do I know |x-a| is less than is"

Does that make sense? That's where i kept losing focus and going round in circles. If you could share some gem of wisdom that would stop me thinking that annoying thought it would be much appreciated

15. Jun 4, 2010

### Staff: Mentor

Somebody gives you epsilon, and says they want |f(x) - f(a)| < epsilon. Your job is to find a delta so that for any x such that |x - a| < delta, then |f(x) - f(a)| < epsilon.

This is especially easy to do for this function, since f(x) = x and f(a) = a. The only simpler problem is f(x) = c, a constant. Any delta will work, no matter how small epsilon is chosen.

16. Jun 4, 2010

### Moogie

Is it because I've got so much freedom i don't know what to do with it?

In other words, it's up to me what epsilon and x and a are. But if i pick a value of epsilon then I can pick a value for x and a such that |x-a| < epsilon. Doesn't matter what they are. But then if delta is less than or equal to epsilon, keeping the same values for x and a, |x-a| < delta <= epsilon?

In other words a and x are arbitary?

17. Jun 4, 2010

### Staff: Mentor

I don't think you get it yet.

a is fixed, and f(a) is fixed. Somebody else challenges you by picking an epsilon. You pick a delta based on that epsilon.

Now, any x you choose has to be within delta of a (i.e., 0 < |x - a| < delta) and if you have chosen delta correctly, f(x) will be within epsilon of L = f(a) (i.e., |f(x) - L| < epsilon.

Think about it in terms of a bow and arrow and the bull's eye on an archery target. Under ideal conditions you could hold the bow just right and the arrow would fly to the exact center of the bull's eye, which is analogous to L.

If you only need to get within, say, a circle of radius epsilon around the exact center, you have some freedom in how you point the arrow when you release the bowstring (analogous to delta). The smaller the target circle, the more precise you have to be on the initial position of the arrow.

18. Jun 4, 2010

### Tedjn

I think you now have the right idea, although I'm not sure what you mean by a and x arbitrary.

Let's try to break this problem down a little more. A function is continuous by definition at a point a if the limit as x tends to a of f(x) is f(a). To prove a function is continuous at all points, you need to prove it for all a. So yes, to do this, we pick an arbitrary a, prove it is true for this a. Convince yourself that this proves it for all a.

Once we have chosen this arbitrary a, we can begin to look at the ε-δ limit definition. We need to show that for all ε > 0, there exists δ > 0, perhaps dependent on ε, such that (insert the rest). To prove this for every ε, we choose an arbitrary ε and prove it for this arbitrary ε. Convince yourself as above that this proves it for all ε.

In the end, what we want to show is that, given some fixed (but arbitrary) a and ε > 0, there exists this δ we have both been talking about such that whenever x is within some δ range of a, f(x) is within some ε range of f(a). In this part, to find δ, thinking in terms of a conversation as Mark44 suggested might help.

Does this clarify things a little more? There are a lot of variables in the limit definition, which can be confusing.

19. Jun 4, 2010

### Red_CCF

Isn't it a bit easier to just apply the definition for continuity? Delta epsilon is a lot more work and it's quite arbitrary/subjective since you can "manipulate" it. I find delta epsilon more useful for proving continuity at a particular given point

20. Jun 4, 2010

### Staff: Mentor

The OP had some questions specifically about delta-epsilon proofs. The definition of continuity at a point is a limit, so underlying it is the whole delta-epsilon business.

What makes it seem arbitrary or subjective?