Calculating $f(0)$ and $f'(0)$ given that $f(2^{-n})>0$ and $f(3^{-n})<0$ Let $f\colon \mathbb{R}\to\mathbb{R}$ be a differentiable function so that $f(2^{-n})>0$ and $f(3^{-n})<0$ for every $n\geq{2}$. Calculate $f(0)$ and $f'(0)$.
I have tried solving it using the Intermediate Value Theorem without any results. How can I solve it? 
 A: Notice that since $f$ is differentiable, it is also continuous.
$2^{-n}$ and $3^{-n}$ both converge to zero.
Using the property you outlined above and continuity, we must have
$f(0)\geq 0$ by taking the limit of $f(2^{-n})$
Similarly, by taking the limit of $f(3^{-n})$, we must have $f(0)\leq 0$.
Thus $f(0)=0$
Notice now that since $f$ is differentiable, the following limits exist
$f'(0)=\lim_{n\to\infty}\frac{f(2^{-n})-f(0)}{2^{-n}-0}$
$f'(0)=\lim_{n\to\infty}\frac{f(3^{-n})-f(0)}{3^{-n}-0}$
But $\frac{f(2^{-n})}{2^{-n}}>0$ for all n, so that $f'(0)\geq 0$
and $\frac{f(3^{-n})}{3^{-n}}<0$ for all n, so that $f'(0)\leq 0$
This of course implies $f'(0)=0$
A: The theorem you want to be using is that if a limit $\lim_{x \to a} g(x)$ exists then the limit may be calculated by sequences. That is, given any sequence $(x_n)$ which converges to $a$ (if $g$ is not continuous at $a$ you also need to ask that $x_n \ne a$ for any $n$),
$$ \lim_{x \to a} g(x) = \lim_{n \to \infty} g(x_n).$$
For your problem, you know that $f$ is differentiable, which means that $f$ is continuous, which means that the limit $\lim_{x \to 0} f(x)$ exists. You also have two sequences, $x_n = 2^{-n}$ and $x_n = 3^{-n}$ which converge. Therefore
$$
\begin{align*}
f(0) &= \lim_{x \to 0} f(x) \qquad \text{ (by continuity)} \\
     &= \lim_{n \to \infty} f(2^{-n}) \quad \text{(by the theorem)} \\
     &= \lim_{n \to \infty} f(3^{-n}) \quad \text{(by the theorem).} \\
\end{align*}
$$
Now you can bring in the inequalities $f(3^{-n}) < 0 < f(2^{-n})$.
A: OUTLINE
Both $\langle 2^{-n} \rangle$ and $\langle 3^{-n} \rangle$ are sequences that go to zero. So, we have $\displaystyle \lim_{x \to 0} f(x) \ge 0$ from the first condition and $\displaystyle \lim_{x \to 0} f(x) \le 0$ from the second condition. Combine with continuity to get:
$$f(0) = \displaystyle \lim_{x \to 0} f(x) = 0$$
Observe that $g(x) := \dfrac{f(x)}x$ is another function satisfying the two conditions. Therefore:
$$f'(0) = \displaystyle \lim_{h \to 0} \dfrac{f(h)-f(0)}{h} = \lim_{h \to 0} g(h) = 0$$

PROOF
For the sake of a contradiction, WLOG assume $\displaystyle \lim_{x \to 0} f(x) = L$ where $L > 0$.
Then, $\forall \epsilon > 0: \exists \delta > 0: \forall x \in \Bbb R: |x| < \delta \implies |f(x)-L| < \epsilon$.
Now, let $\epsilon = \dfrac L2$, and the corresponding $\delta$ be $\delta_0$.
Pick $x = 3^{-\lceil -\log_3 \delta_0 \rceil}$, whose existence is guaranteed by the Archimedean Principle.
We see that $|x| = 3^{-\lceil -\log_3 \delta_0 \rceil} < 3^{\log_3 \delta_0} = \delta_0$.
However, $f(x)<0$ from the condition, which contradicts $|f(x)-L| < \epsilon$.
