# 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.

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Could you be more specific about your statement : "this inequality would not hold for a smaller $k$" ? Does this mean $$\forall C'>0, \ \forall k'<k, \ \exists x>0, \ |f(x)| > C'/x^{k'}$$ ? – vanna Sep 24 '12 at 13:29
Yes, that's what I mean, vanna. – Homer Sep 24 '12 at 14:16
The whole elaborated setup is just a terribly fancy way to say that $k=0$. Where did the problem come from? – fedja Oct 31 '12 at 12:29
Is $k$ assumed to be integer? And is the claim that the limit is finite, or would $+\infty$ or $-\infty$ also be included in "the limit exists"? – Lukas Geyer Oct 31 '12 at 23:05

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.
You don't says what is $C_3$ anywhere – leo Nov 1 '12 at 4:34
It is just some constant, you can choose $C_3=C_1+C_2$, if you want an explicit expression. – Lukas Geyer Nov 1 '12 at 4:43