Verified answer

I think you mean the n derivative of f not (f(x))^n, right?

If so, you have one smooth function because it is not only analytic but every derivative is Lipschitz everywhere. That's a pretty limited class of functions. But you can surely say that such a function is equal to its Taylor series expansion, so the limit there = 0.

Edit: Albswok's proof is good, but you need way less than every derivative bounded by a finite constant everywhere for a function to equal its Taylor series expansion.

That first guy defined continuous functions, which is silly because analytic and wicked smooth is much stronger than continuous.

Suppose we have a function that f is a function such that f^n(x) exists for all n and such that |f^n(x)|≤M for some M. If Tn(x) represents the Taylor polynomial for f centered at a=0, from above. Such a function can be represented by a graph in the cartesian plane; the function is continuous if, roughly speaking, the graph is a single unbroken curve with no "holes" or "jumps": if it can be drawn by hand without lifting the pencil from the paper.

To be more precise, we say that the function f is continuous at some point c if the following three requirements are satisfied:

* f(c) must be defined (i.e. c must be an element of the domain of f)

* The limit of f(x), as x approaches c, must exist

* The limit of f(x), as x approaches c, must equal f(c)

We call the function everywhere continuous, or simply continuous, if it is continuous at every point of its domain.

Epsilon-delta definition

Without resorting to limits, one can define continuity of real functions as follows.

Again consider a function f that maps a set of real numbers to another set of real numbers, and suppose c is an element of the domain of f. The function f is said to be continuous at the point c if (and only if) the following holds: For any positive number ε however small, there exists some positive number δ such that for all x with c - δ < x < c + δ, the value of f(x) will satisfy f(c) - ε < f(x) < f(c) + ε. This "epsilon-delta definition" of continuity was first given by Cauchy.

More intuitively, we can say that if we want to get all the f(x) values to stay in some small neighborhood around f(c), we simply need to choose a small enough neighborhood for the x values around c, and we can do that no matter how small the f(x) neighborhood is.

Well, Taylor's theorem tells us that for any n times differentiable function, and if T_n is the n - 1 degree taylor polynomial centred at x = a, then for all x:

f(x) - T_n(x) = f^(n)(c) / n! * (x - a)^n

for some c between a and x.

But, if M is such that |f^(n)(x)| <= M for all n and x, then this tells us that:

|f(x) - T_n(x)| <= M / n! * (x - a)^n

So, what happens to this limit as n approaches infinity? Well, we have an exponential over a factorial, multiplied by the constant M. As such, it will go to 0 (although this may require proof, if you haven't proved it before). Since |f(x) - T_n(x)| is also bounded below by 0, |f(x) - T_n(x)| is bounded by two 0-convergent sequences, so by squeeze theorem, it too must converge to 0, which implies that:

f(x) - T_n(x) ---> 0