$$f(n+1)-f(n)=\frac{f(n+1)-f(n)}{(n+1)-n}=f'(c_n)\;\;,\;\;\;c_n\in (n,\,n+1)$$
using the MVT (Lagrange's Theorem) for differentiable functions.
From this it follows that $\;c_n\xrightarrow[n\to\infty]{}\infty\;$, and thus
$$\alpha=\lim_{n\to\infty}f'(c_n)=\lim_{n\to\infty}\left(f(n+1)-f(n)\right)=1-1=0$$
Refer from following link:
https://math.stackexchange.com/a/1257166/952652