Let $f:R\rightarrow R$ be a non constant, three times differentiable function. If $f(1+\frac{1}{n})=1$ for all integers n,then f'(1) =? Let $f:R\rightarrow R$ be a non  constant, three times differentiable function. If $f(1+\frac{1}{n})=1$ for all integers n,then $f'(1) = ?$
I was trying hard to find a function which satisfies but it didnot work. Help will be appreciated.Thanks
 A: By the Mean Value Theorem, for every positive integer $n$ there is a $c_n$ between $1+\frac{1}{n+1}$ and $1+\frac{1}{n}$ such that $f'(c_n)=0$. It follows by the continuity of $f'$ that $f'(1)=0$.
Remark: Note that by repeating the argument with $c_n$ instead of $1+\frac{1}{n}$, we can also conclude that $f''(1)=0$.
But one can go a little further, and conclude there is a decreasing sequence  $d_n$ of points, with limit $1$, such that $f''(d_n)=0$.  We have $\frac{f''(d_n)-f''(1)}{d_n}=0$ for all $n$. Since $f''$ is differentiable, we conclude that $f'''(1)=0$.
The reason we computed $f'''(1)$ is that the condition $f$ is three times differentiable is far stronger than what is needed to show that $f'(1)=0$. So one might as well squeeze the maximum possible amount of information from the situation.
A: If $f(1+1/n) = 1$
for all $n$,
then,
since $f$ is differentiable,
it is continuous,
so $f(1) = 1$.
Then
$0
=\frac{f(1+1/n)-1}{1/n}
=\frac{f(1+1/n)-f(1)}{1/n}
=f'(1)
$
from the definition
$f'(x)
=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}
$
with
$x = 1$
and
$h = \frac1{n}$.
A: The fact that $f$ is three times differentiable, suggests using Taylor developments; let's try with Taylor at degree $1$:
$$
f(1+h)=f(1)+hf'(1)+hR_1(h)
$$
where $\lim_{h\to0}R_1(h)=0$. If we apply this for $h=1/n$, we get
$$
1=f\left(1+\frac{1}{n}\right)=
f(1)+\frac{1}{n}f'(1)+\frac{1}{n}R_1\left(\frac{1}{n}\right)
$$
Letting $n\to\infty$, we get $f(1)=1$. Thus we can rewrite the relation as
$$
0=f'(1)+R_1\left(\frac{1}{n}\right)
$$
and again, letting $n\to\infty$, we conclude $f'(1)=0$.
Now we can use Taylor at degree $2$:
$$
f(1+h)=1+\frac{h^2}{2}f''(1)+h^2R_2(h)
$$
For $h=1/n$ and multiplying by $n^2$ before doing the limit as $n\to\infty$, we deduce $f''(1)=0$. Similarly, $f'''(1)=0$.
(Note: it's just the same argument as André Nicolas', just from a different point of view.)
How can we find an example? If we set $g(x)=f(x+1)-1$, we have $g(0)=g'(0)=0$ and $g(1/n)=0$ for all $n$. Consider $G(x)=g(1/x)$ (for $x\ne0$): $G(n)=0$. Then any (three times) differentiable function multiplied by $\sin(1/(\pi x))$ would do. We just need to ensure three times differentiability at $0$:
$$
g(x)=x^4k(x)\sin\frac{1}{\pi x}
$$
where $k$ is any three times differentiable function.
A: $f$ is continuous, since it is differentiable. So $f(1)=1= \lim_n f(1+1/n)$
A: Following on from @egreg's answer, you could generate an even larger family of example functions by finding any $3$-times differentiable function, $r$, which equals $0$ at the reciprocal integers: i.e. $r(x)=0$ for $\frac1{x}\in\mathbb{Z}$. Then, $F(x)=k(x)r(x)$ where $k$ is any $3$-times differentiable function, will give you a $f(x)=F(x-1)+1$ that satisfies the constraints.
Why must $r$ be $3$-times differentiable? We can easily show that $f'''(x)=F'''(x-1)$ so if $F'''$ exists, so does $f'''$. But:
$$\begin{aligned}
F'''(x)=&k'''r+k''r'\\
&2(k''r'+k'r'')\\
&k'r''+\color{red}{kr'''}\\
\end{aligned}$$
Due to the presence of the $r'''$ term (in red), we want our $r$ to be $3$-times differentiable, else we won't have an $f'''$. We also can't let $k=0$ since we want a function that's non-constant.
We can pick any number of $3$-times differentiable functions which equal $0$ at the reciprocal integers, for instance $r(x)=x^p\sum_{n=1} a_n \sin(\frac{n\pi}{x})$, where $a_n$ are arbitrary constants and where $p\geq4$.
The graph below shows once such $r$, where $p=5$ and $(a_n)=(1,2,0,-0.5\ldots)$.

This could give you a possible $f(x)=k(x)\,(x-1)^5\left[ \sin(\frac{\pi}{x-1})+ 2\sin(\frac{2\pi}{x-1})-0.5 \sin(\frac{4\pi}{x-1})\right]+1$, or many other functions which fit the bill.
