Show that a limit exists Let $f : (0, \infty) \rightarrow \mathbb{R}$ be differentiable and suppose that $|f(x)| \leq \frac{C}{x^k}$ ($k$ is non-negative) and this inequality would not hold for a smaller $k$ (even if you change $C$). Suppose this also holds for $|f'(x)|$ but with a possibly different $C$, but same $k$. Show that $\lim_{x \rightarrow 0^+} f(x)$ exists.
 A: Taking the maximum of the constants for $f$ and $f'$ we can assume that $$|f(x)| \le C x^{-k} \text{ and } |f'(x)| \le Cx^{-k} \text{ for all } x>0.$$ Integrating the inequality for $f'$ gives $$|f(x)| \le \begin{cases} C_1 + C_2 x^{-(k-1)} & \text{for } k \ne 1 \\ C_1 + C_2 |\ln x| & \text{for } k=1\end{cases}$$
with some constants $C_1$ and $C_2$.
It is still not absolutely clear to me whether $k$ is supposed to be integer, and whether an infinite limit is allowed or not, but at least this estimate shows that the assumption is never satisfied for $k>1$, as follows. For $k>1$ the estimates imply that there exists $C_3$ with $$ |f(x)| \le C_3 x^{-(k-1)} \text{ for } 0<x<1 $$ and $$ |f(x)| \le C x^{-k} \le C x^{-(k-1)} \text{ for } x \ge 1.$$ Setting $C_4 = \max (C,C_3)$ we get $$ |f(x)| \le C_4 x^{-(k-1)} \text{ for } x>0. $$ This contradicts the assumption that this inequality does not hold for exponents $k'<k$.
Obviously the estimate can be satisfied for $k=0$, and in that case boundedness of the derivative implies that $f$ is uniformly Lipschitz continuous, hence it extends continuously to $[0,\infty)$.
For the case $0 < k \le 1$ I am waiting for clarification of the exact statement of the problem.
