# [f(x)'^n continuous at a point (EXAM IN 4 hrs.)

• llursweetiell
In summary, the conversation discusses how to use the epsilon-delta definition of continuity to prove that a function g(x)=[f(x)]^n is continuous at a, given that f:D-->R is continuous at a and n > 1 is a positive integer. The conversation includes a previous attempt using the composition proof, but ultimately the participants discuss the use of the epsilon-delta definition to estimate the difference between [f(x)]^n and [f(a)]^n. They conclude that by choosing an appropriate delta, the difference can be made less than epsilon, thus proving the continuity of g(x) at a.
llursweetiell

## Homework Statement

Suppose f: D-->R is continuous at a. Let n >1 be a positive integer. using the epsilon-delta definition of continuity, prove g(x)=[f(x)]^n is continuous as a

## Homework Equations

i know how to do it as a sequence proof; but i don't know how to use the epislon/delta definition to prove it.

## The Attempt at a Solution

i tried using the composition proof too, but that didn't work. any help?

Last edited:
Um, I'm assuming you mean g = f nested n times since you mentioned the "composition proof". If this is the case, I don't even think the theorem is true since if n = 2, you would need f to be continuous at f(a), but f(a) might not even be in D.

the problem states:
suppose f:D--> R is continuous at a. Let n >1 be a positive integer. Using the epsilon and delta definition of continuity, prove g(x)=[f(x)]^n is continuous at a.

Does that help?

Also, the composition way did not work. which is why I'm unsure of how to go about tackling the problem.

Last edited:
Sorry, the comment about the composition of functions threw me off. I should have known that the solvable interpretation is the right one.

Anyways, you need to estimate |[f(x)]^n - [f(a)]^n| knowing that you can bind |f(x) - f(a)| by any positive number. But a^n - b^n has a very nice and symmetric formal factorization, namely

$$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + b^{n-2}a + b^{n-1})$$

Thus you can replace a with f(x) and b with f(a) to get |[f(x)]^n - [f(a)]^n| equal to |f(x) - f(a)| times a bunch of terms within an absolute value (this is key) as a result of the factorization above. But now can you see what to bind |f(x) - f(a)| by to ensure that |[f(x)]^n - [f(a)]^n| is less than say, epsilon?

So i wrote,
We know that f is continuous at a. to prove that g is continuous at a:
let epsilon be greater than 0 and let delta > 0 such that d=epsilon/$$(f(x)^{n-1} + f(x)^{n-2}f(a) + ... + f(a)^{n-2}f(x) + f(a)^{n-1})$$

then,
for |x-a| < delta, then |f(x)^n-f(a)^n|=$$f(x)^n - f(a)^n = (f(x)-f(a))(f(x)^{n-1} + f(x)^{n-2}f(a) + ... + f(a)^{n-2}f(x) + f(a)^{n-1})$$
< (f(x)-f(a)) * delta = epsilon. So g is continuous at a.

Is that what you meant?

Last edited:
You have the right idea, but you don't actually get to choose delta here. Since f is continuous at a, there is already some delta for which |x-a| < delta implies
$$|f(x) - f(a)| < \frac{\varepsilon}{(f(x)^{n-1} + f(x)^{n-2}f(a) + ... + f(a)^{n-2}f(x) + f(a)^{n-1})}.$$

But this is exactly what we need.

[tex said:
|f(x) - f(a)| < \frac{\varepsilon}{(f(x)^{n-1} + f(x)^{n-2}f(a) + ... + f(a)^{n-2}f(x) + f(a)^{n-1})}.[/tex]

QUOTE]

do you mean f(x)^n-f(a)^n < all of the factorization?

Eh, perhaps it would be clearer if I just pieced together the proof. Let epsilon > 0 be given. By the continuity of f at a, there exists some delta > 0 such that |x-a| < delta implies
$$|f(x) - f(a)| < \frac{\varepsilon}{|f(x)^{n-1} + f(x)^{n-2}f(a) + ... + f(a)^{n-2}f(x) + f(a)^{n-1}|}.$$
But then
$$|f(x)^n - f(a)^n| = |f(x)-f(a)||f(x)^{n-1} + f(x)^{n-2}f(a) + ... + f(a)^{n-2}f(x) + f(a)^{n-1}| < \varepsilon.$$

oh that makes sense. thank you so much! =)

## 1. What does it mean for a function to be continuous at a point?

Continuity at a point means that the value of the function at that point is equal to the limit of the function as the input approaches that point.

## 2. How is continuity at a point determined?

To determine continuity at a point, we need to check if the limit of the function exists at that point and if it is equal to the value of the function at that point. If both conditions are met, the function is continuous at that point.

## 3. What is the significance of continuity at a point?

Continuity at a point is important because it ensures that there are no abrupt changes or discontinuities in the function at that point. This makes the function more predictable and easier to analyze.

## 4. Can a function be continuous at a point but not on its entire domain?

Yes, it is possible for a function to be continuous at a point but not on its entire domain. This can occur if the function has a jump or a removable discontinuity at a certain point.

## 5. How can we prove that a function is continuous at a point?

To prove that a function is continuous at a point, we can use the epsilon-delta definition of continuity. This involves showing that for any small value of epsilon, we can find a corresponding value of delta such that the distance between the input and the point in question is less than delta, the corresponding output is within epsilon of the value of the function at that point.

Replies
22
Views
805
Replies
6
Views
790
Replies
7
Views
429
Replies
14
Views
805
Replies
2
Views
610
Replies
13
Views
1K
Replies
7
Views
2K
Replies
26
Views
2K
Replies
3
Views
1K
Replies
2
Views
1K