Consider a sequence $x_{k + 1} = f(x_{k})$, where $f \in C^{3}$. Moreover, assume $f(0) = f'(0) = f''(0) = 0$. Show that there exists a continuous function $g$ such that
$$|x_{k+1}| \leq g(x_{k})|x_{k}|^{3}.$$
I tried rewriting it as a function in $f$, but I didn't get anywhere. I got
$$|f(x_{k})|/|x_{k}|^{3} \leq g(x_{k}), $$
from which I have been stuck. I'm guessing that you need to use the fact that $f^{(n)} = 0$ for $n = 1, 2$ since they wouldn't have given it to us otherwise?
Let's define $$g(x) = \left|\frac{f(x)}{x^3}\right| \text{ if } x\neq 0 \quad \text{ and } \quad g(0)=\frac{f'''(0)}{6}$$
Obviously, you get for all $k$ that $$|x_{k+1}| = g(x_k) |x_k|^3$$
Now you have to prove that $g$ is continuous. It is the case on $\mathbb{R} \setminus \lbrace 0 \rbrace$ ; moreover, Taylor Lagrange formula between $0$ and $x$ shows that for all $x$, there exists $z_x \in [0,x]$ such that $$f(x)=x^3\frac{f'''(z_x)}{6}$$
So for all $x \neq 0$, $$|g(x)-g(0)|=\left|\frac{f'''(z_x)-f'''(0)}{6}\right|$$
which tends to $0$, because $z_x$ tends to $0$ as $x$ tends to $0$, and $f'''$ is continuous.
This shows that $g$ is continuous also in $0$.