What does $\lim\limits_{x \to \infty} f(x) = 1$ say about $\lim\limits_{x \to \infty} f'(x)$?

Nothing. you can take something like: $x\mapsto \frac{\sin(x^2)}{x}+1$.

This is in response to Balaji's comment.

Lemma. Let $f\colon (a,+\infty)\to\mathbb{R}$ be a bounded $C^1$ function such that $\lim_{x\to+\infty}f'(x)=\delta$ exists. Then $\delta=0$.

Proof. We proceed by contradiction and assume that $\delta\ne0$. Suppose first that $\delta>0$. Then there exists $x_0>a$ such that $f'(x)\ge\delta/2$ for all $x\ge x_0$. Then $$ f(x)=f(x_0)+\int_{x_0}^xf'(t)dt\ge f(x_0)+\frac{\delta}{2}(x-x_0)\quad\forall x\ge x_0. $$ In particular $f$ is unbounded. The case $\delta<0$ is done similarly.

If you do not want to use integrals, you may reason as follows. Let $$g(x)=f(x)-f(x_0)-\frac{\delta}{2}(x-x_0).$$ Then $g(x_0)=0$ and $g'(x)\ge 0$ for all $x>x_0$. It follows that $g$ is increasing in $(x_0,+\infty)$ and $g(x)\ge 0$ for all $x>x_0$.