1. The problem statement, all variables and given/known data 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 2. Relevant 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. 3. The attempt at a solution i tried using the composition proof too, but that didn't work. any help?
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.
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 [tex]a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + b^{n-2}a + b^{n-1})[/tex] 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/[tex](f(x)^{n-1} + f(x)^{n-2}f(a) + ... + f(a)^{n-2}f(x) + f(a)^{n-1})[/tex] then, for |x-a| < delta, then |f(x)^n-f(a)^n|=[tex]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})[/tex] < (f(x)-f(a)) * delta = epsilon. So g is continuous at a. Is that what you meant?
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 [tex]|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] But this is exactly what we need.
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 [tex] |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] But then [tex] |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. [/tex]