<s>[INST] Rate of convergence for sequences I have numerically determined that the sequence $\{f_x\} = \frac{\sin(x^2)}{x^2}$ approaches $1$ (as $x$ approaches $0$) faster than the sequence $\{g_x\} = \frac{\sin^2(x)}{x^2}$. However, I am stuck for determing the rate of convergence of each sequence.
Should I use the small angle approximation $\sin(x)\approx x$ and the fact that $|\sin(x)|\leq 1$? Thanks.
 [/INST] Let's start with the definition of rate of convergence of functions (There is also definition of rate of convergence of sequences):

Definition. Let $f$ be a function defined on the interval $(a,b)$ that
  contains $x=0$, and suppose $\lim_{x\to 0}{f(x)=L}$. If there exists a
  function $g$ for which $\lim_{x\to 0}{g(x)=0}$ and a positive constant
  $K$ such that $$|f(x)-L| \le K|g(x)|$$ for all sufficiently small
  values of $x$, then $f(x)$ is said to converge to $L$ with rate of
  convergence $O(g(x))$.

Let $f(x)=\sin{x^2}/{x^2}$ and $g(x)=(\sin{x})^2/{x^2}$. From Taylor's Theorem, we know that
$$\sin{x}=x-\frac{1}{6}x^{3}\cos{\xi}$$
for some $\xi$ between $0$ and $x$. Hence,
$$f(x)=\frac{x^2-\frac{1}{6}x^{6}\cos{\xi}}{x^2}=1-\frac{1}{6}x^{4}\cos{\xi}$$
We know $\sin{x}=(1-\cos{2x})/{2}$ and
$$\cos{x}=1-\frac{1}{2}x^2+\frac{1}{24}x^4\cos{\xi}$$
for some $\xi$ between $0$ and $x$. Therefore,
$$g(x)=\frac{(1-1+2x^2-\frac{2}{3}x^4\cos{\xi})/2}{x^2}=\frac{x^2-\frac{1}{3}x^4\cos{\xi}}{x^2}=1-\frac{1}{3}x^2\cos{\xi}$$
Finally
$$|\frac{\sin{x^2}}{x^2}-1|=\frac{1}{6}|x^{4}\cos{\xi}|\le \frac{1}{6}|x^{4}|$$
and
$$|\frac{(\sin{x})^2}{x^2}-1|=\frac{1}{3}|x^2\cos{\xi}|\le \frac{1}{3}|x^2|$$
It follows that the rate of convergence of $f(x)$ is $O(x^4)$ and the rate of convergence of $g(x)$ is $O(x^2)$. It is common to express this in shorthand by writing
$$\frac{\sin{x^2}}{x^2}=1+O(x^4) \qquad and \qquad \frac{(\sin{x})^2}{x^2}=1+O(x^2)$$
That means an error term with rate of convergence $O(x^4)$ approaches zero faster than an error term with rate of convergence $O(x^2)$, as the value of $x$ approaches zero. Therefore $f(x)$ approaches one faster than $g(x)$ as $x$ approaches zero.
Source: A Friendly Introduction to Numerical Analysis by Brian Bradie.
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