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

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.
