Notation Clarification in Taylor Expansion I'm reading a book that states the following:

Suppose we have a continuously differentible function $f(t)$. Around each $t_0$ we can expand $f(t)$,
  $$f(t) = f(t_0) + f'(t_0)(t-t_0) + o(t-t_0)$$

Shouldn't the last term read $o\big((t-t_0)^2\big)$? Initially I thought it was a mistake, but they kept writing it, so now I think there's something I'm missing.

No. $f(x)=o(g(x))$ as $x \to x_0$ if $f(x)/g(x) \to 0$ as $x \to x_0$. Indeed, the equation the book gives is equivalent to the definition of the derivative: $f$ is differentiable at $t_0$ if and only if there is a number $f'(t_0)$ so that
$$ \frac{f(t)-f(t_0)}{t-t_0} - f'(t_0) \to 0 $$
as $t \to t_0$, which is the same as saying that $f(t) = f(t_0) + f'(t_0)(t-t_0) + o(t-t_0)$.
On the other hand, $f(x)=O(g(x))$ as $x \to x_0$ just means that $ \lvert f(x) \rvert \leq C\lvert g(x) \rvert$ for some fixed $C$ for $x$ sufficiently close to $x_0$.
The function $f(x)=x^{4/3}$ provides an example that shows that continuously differentiable does not give you $O(x^2)$: we have $f(0)=f'(0)=0$, but it is certainly not true that $x^{4/3} \leq C x^2$ for some fixed $C$ as $x \to 0$ since $ x^{4/3-2} = x^{-2/3} \to \infty$ as $x \to 0$.
