I decided to put my attempt at a solution before the question, because the "solution" is what my question is about.(adsbygoogle = window.adsbygoogle || []).push({});

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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**