Let $f:\mathbb{R} \rightarrow \mathbb{R} $ be a non-constant, three times differentiable function. If $f(1+\frac{1}{n})=1$ for all integers $n$, then find $f''(1)$.
I could find $f'(1)=\lim_{n \rightarrow \infty} \frac{f(1+\frac{1}{n})-f(1)}{\frac{1}{n}}$. Now, $f(1)=1$ due to continuity of $f$ at $1$. But the expression of $f''(1)$ is becoming overtly complicated. Please help.
Observe that since we know that $\;f''(1)\;$ exists, for $\;n\in\Bbb N\;$ , we have that (use for example Check my workings: Show that $\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$)
$$f''(1)=\lim_{n\to\infty}\frac{f\left(1+\frac1n\right)-2f(1)+f\left(1-\frac1n\right)}{\left(\frac1n\right)^2}=\lim_{n\to\infty}\frac{1-2+1}{\frac1{n^2}}=0$$
Because, as you say, by continuity $\;f(1)=1\;$ .
Warning: One equality sign before the last one I am not passing to the limit in the numerator while keeping the denominator as it is. This would require justification. What I'm doing is substituting the actual values of $\;f\;$ at those points!