Why is $f(x+h)=f(x)+hf'(x)+O(h^2)$? I'm reading Chapter 3 of Introduction to Calculus and Analysis by Richard Courant and Fritz John. On page 257 they give the equations
$$ f(x+h) - f(x) = hf'(x) + o(h), $$ and
$$ f(x+h) - f(x) = hf'(x) + O(h^2) $$
for $h \to 0,$
with the definitions $f=o(g)$ if $\lim f(x)/g(x) = 0$ and $f=O(g)$ if the ratio $f(x)/g(x)$ is bounded.
Also, the hypotheses on $f$ for the first equation are $f$ is differentiable at x, and for the second equation that $f''$ is defined and continuous.
I can see why the first equation is true, because
$$ \lim_{h \to 0}\frac{f(x+h)-f(x)-hf'(x)}{h}=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} - f'(x)=0. $$
But for the second equation, how do I show that $$ \frac{f(x+h) - f(x) -hf'(x)}{h^2} $$ is bounded?
 A: Note that there is a form of Taylor's theorem with very minimal constraints namely:
Taylor's Theorem with Peano's Form of Remainder: Let $f$ be a function such that $f^{(n)}(a)$ exists. Then we have $$f(a + h) = f(a) + hf'(a) + \frac{h^{2}}{2!}f''(a) + \cdots + \frac{h^{n}}{n!}f^{(n)}(a) + o(h^{n})$$
For $n = 1$ (which is also the first result in your question with $x$ in place of $a$) this is an immediate result of definition of derivative $f'(a)$ and hence it does not require any proof. The proof for $n > 1$ follows by the application of L'Hospital's Rule applied $(n - 1)$ times.
Your second result requires some more conditions on $f$. Let's see why. The only difference between your first and second result is replacing $o(h)$ by $O(h^{2})$. This is not true in general because we have $F(h) = h^{3/2} = o(h)$ and clearly if we divide by $h^{2}$ we get $F(h)/h^{2} = h^{-1/2}$ which is not bounded as $h \to 0$. So we can't replace $o(h)$ by $O(h^{2})$ in general. 
Let's now see what happens when we put $n = 2$ in Taylor's theorem above. We get $$f(a + h) - f(a) = hf'(a) + h^{2}f''(a)/2 + o(h^{2})$$ which implies that $$f(a + h) - f(a) = hf'(a) + O(h^{2})$$ Thus for you second result we need the existence of $f''$ at the point under consideration. You can see that we don't need the continuity of $f''$ at the point under consideration.

I can see from your comments (to other answers here) that you are not familiar with Taylor's Theorem. In case you are familiar with L'Hospital's Rule (meaning you know the conditions under which it can be applied and perhaps its slightly non-trivial proof) I can show that your second result holds if $f''$ exists (no need of continuity).
We have
\begin{align}
L &= \lim_{h \to 0}\frac{f(a + h) - f(a) - hf'(a)}{h^{2}}\notag\\
&= \lim_{h \to 0}\frac{f'(a + h) - f'(a)}{2h}\text{ (applying L'Hospital's Rule)}\notag\\
&= \frac{f''(a)}{2}\notag
\end{align}
It follows from the existence of above limit that the ratio $$\frac{f(a + h) - f(a) - hf'(a)}{h^{2}}$$ is bounded as $h \to 0$ and this is what we mean by saying $$f(a + h) - f(a) - hf'(a) = O(h^{2})$$

Here is perhaps what might be more satisfactory for you in case you don't know L'Hospital's Rule. But then we need the existence as well as continuity of $f''$ at point $a$. Let's see how. By the Mean Value Theorem (you have to know it otherwise this question is not at your level) we have $$f(a + h) - f(a) = hf'(c)\tag{1}$$ where $c$ is some number between $a$ and $a + h$. Now applying mean value theorem again we can see that $$f'(c) - f'(a) = (c - a)f''(d)\tag{2}$$ where $d$ lies between $a$ and $c$ and therefore between $a$ and $a + h$. Thus we have via equations $(1)$ and $(2)$ above $$f(a + h) - f(a) = h\{f'(a) + (c - a)f''(d)\} = hf'(a) + h(c - a)f''(d)$$ Since $f''$ is continuous at $a$ it follows that $f''$ is bounded in neighborhood of $a$ and hence $f''(d)$ is bounded. And clearly $|h(c - a)| \leq h^{2}$ and thus we can see that $$f(a + h) - f(a) = hf'(a) + O(h^{2})$$
A: Note that if $f(x)$ is twice differentiable, then the second part will be immediately from  Tayor theorem. If not, it need not be true. 
Or using Langerage meanvalue theorem, we obtain
$$f(x+h)=f(x)+f'(x)h+\frac{f''(\xi)}{2}h^2.$$
