Limit of the derivative of a function

By Mean Value theorem we have the following equation $$f(x + 1) - f(x) = f'(\xi)$$ where $x < \xi < x + 1$. The LHS of the above equation tends to $1 - 1 = 0$ as $x \to \infty$ and RHS tends to $s$ as $x \to \infty$. Hence $s$ must be $0$.

Since $$\lim_{x\to\infty}f'(x)=s$$

That means for all $\epsilon>0$, there exists $R$ and $\delta>0$ s.t. $$\bigg|\frac{f(x+h)-f(x)}{h}-s\bigg|<\epsilon$$ whenever $x>R$ and $h<\delta$.

In other words, $$s-\epsilon<\frac{f(x+h)-f(x)}{h}<s+\epsilon$$

Now suppose $s\ne 0$, then set $\epsilon<|s|/2$ and $h=\delta/2$, we have $$\bigg|f(x+\delta/2)-f(x)\bigg|>\frac{|s|\delta}{4}$$

However, since $$\lim_{x\to\infty}f(x)=1$$ we know that there exist $R'$ so that $$\bigg|f(x)-1\bigg|<\frac{|s|\delta}{8}$$ whenever $x>R'$.

Now for $x>\max(R,R')$, we have $$\bigg|f(x+\delta/2)-f(x)\bigg| \le |f(x+\delta/2)-1|+|f(x)-1|<\frac{|s|\delta}{4}$$ which is a contradiction.

Hence $s=0$.

If the derivative does not approach zero at infinity, the function value will continue to change (non-zero slope). Since we know the function is a constant, the derivative must go to zero.

Just pick an $|s| < 1,$ and draw what happens as you do down the real line. If $s \neq 0,$ the function can't remain a constant.