I am a little stuck when it comes to finding the rate of convergence for the function $f(x)=\frac{1-cos(x)}{x^2}$. The first part of this question was to show that the limit was $\frac{1}{2}$, and I was able to accomplish that using the Maclaurin series representation since my professor said we needed that to find the rate of convergence.
The line right before I evaluate the limit is as follows: $\lim{x \to 0} (f(x))=\lim{x \to 0} (\frac{1}{2}-\frac{x^2}{4!}+\frac{x^4}{6!}-...)=\frac{1}{2}$.
Now, I haven't attended the lecture of finding the rate of convergence yet, but that is because I am not able to attend that specific lesson. My professor tried to explain it to me, and by what he said I believe that the rate of convergence for this specific $f(x)$ is $2$.
He was able to just look at the last line of a limit of a function that was evaluated using the Maclaurin series, and state the rate of convergence.
When I looked in the book and researched online, it looked a lot more complicated than just looking at it and knowing what it was.
So, my question is, am I correct in saying that the rate of convergence of $f(x)$ is $2$?
What I think you're referring to is the order. This is the number $p$ with the property that $\frac{f(x+h)-f(x)}{h^p}$ neither goes to zero nor blows up as $h \to 0$. This is indeed $2$ in your case because the second term of the Maclaurin series is a $x^2$ term (since $\cos'(0)=0$).
The order alone doesn't quite quantify the rate of convergence; compare, say, the rate at which $x$ and $100x$ go to zero as $x \to 0$.