Rate of Convergence for sin(1/x^2) with Maclaurin is undefined?

  1. I decided to put my attempt at a solution before the question, because the "solution" is what my question is about.

    1. The problem statement, all variables and given/known data
    Find the rate of convergence for the following as n->infinity:
    lim [sin(1/n^2)]
    n->inf

    Let f(n) = sin(1/n^2) for simplicity.

    2. The attempt at a solution
    I was searching through other forums and resources, and finally found a solution.
    It said to use the Maclaurin Series (thus x0 = 0), but this would make every term
    to look like:

    f(0) + f'(0)*(n^1) + (1/2)*f''(0)*(n^2) + (1/4)*f'''(0)*(n^3) + ... + remainder

    3. Relevant equations
    How can we solve when 1/0 is undefined?
    For example, in *every* term we have a sin(1/0^x) somewhere. This doesn't work, obviously.

    This is solvable using the first few terms of the Maclaurin polynomial according to other sources, but I do not understand how. Did I overlook something? Or am I fundamentally misunderstanding the question?

    Thanks!
     
  2. jcsd
  3. Oh dear. I just figured it out.

    We shouldn't be finding the Maclaurin series for sin(1/x^2) at all. It should be for sin(x) first (with x0=0), then we let x=1/x^2 afterwards. Silly me...
     
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