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

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
$h = \frac1{n}$.
