Differentiation of a function Let $f:\mathbb R \rightarrow \mathbb R$ be a non-constant, thrice  differentiable function such that $f(1+1/n)=1$ for all $n \in \mathbb Z$.  then what is the value of $f^{\prime \prime}(1)$? 
Answer is 0
 A: Suppose we have a two times continuously differentiable function $g$ such that $g''(1)>0$, then we can find an interval $[1,1+\delta)$ such that $g''(x)>0$ for all $x$ in the interval.
The function $g$ cannot take on any given value more than $3$ times in this interval, why?
It is a consequence of the mean value thoerem, suppose that $a_0<a_1<a_2$ and $g(a_j)=z$ for all values.
Then by Rolle's theorem there are two points $a_0<b_1<a_1<b_2<a_2$ with $g'(b_0)=g'(b_1)=0$. A contradiction, since $g''$ is positive in $[1,1+\delta)$, which implies $g'$ is increasing in the same interval.
The case when $g''(1)<0$ is analogous.
Can you see why this result solves your problem? It is because no matter what $\delta$ you obtain you get an infinite number of values in $[1,1+\delta)$ that map to $1$. So we must have $g''(1)=0$.
A: Since $f$ is continuous at $x=1$ we have $f(1) = \lim\limits_{n\to\infty}f(1+1/n) = 1$. We can now use the symmetric formulation for the second derivative to get
$$f''(1) = \lim_{h\to0} \frac{f(1+h)+f(1-h)-2f(1)}{h^2} = \lim_{n\to\infty} \frac{f(1+1/n) + f(1-1/n) - 2f(1)}{1/n^2} = 0$$
since $f(1+1/n) + f(1-1/n) - 2f(1) = 1 + 1 - 2 \equiv 0$ for all $n\in\mathbb{N}$.
A: Hint: we have  $f (t)=1$ for  all  $t \ne 1$.
