Homework problem here, would appreciate an explanation to the answer of this question.
Problem: Find the rate of convergence of
$$ \lim\limits_{h \to 0} \frac{\sin(h)}{h} = 0 $$
The book solves this by using the expansion of the Taylor series. But only too the second polynomial. Like so:
$$ \left|\frac{\sin(h)}{h} -1 \right| = \left| \frac{h – \frac{h^3}{6}\sin(x(h))}{h} – 1 \right| = \left|-\frac{h^2}{6}\sin(x(h)) \right| = O(h^2) $$
(Where $x(h) \in (0,h)$)
But im not sure how they arbitrarily just chose to expand to the second taylor polynomial. What gave them the intuition to use the Taylor Polymnomial in the first place? More to the point, what did they choose to expand the polynomial about?
If anyone could help me that would be great. thanks!
Best Answer
A brief answer to all points:
They chose to expand about $0$ because the limit was as $h \to 0$. That tends to work out best in most cases.
The key to choosing how far to expand is to simply expand "enough". In fact, you could expand further if you want to, but you'd find that the same basic steps apply. The only difference is that we'd be left with, for example, $$ \cdots = \left|-\frac 16 h^2 + \frac 1{120} h^4\right| = o(h^2) $$ In general, the Taylor polynomial centered at $x = a$ is a good way of dissecting how a function "behaves" near $a$, and is generally useful in limit problems like these.