Given that $f$ is differentiable, what does $\lim\limits_{x \to \infty} f(x) = 1$ say about $\lim\limits_{x \to \infty} f^\prime(x)$ ? Intuitively I feel that it's $0$.
I attempted to solve this by trying to evaluate $\lim_{x \to \infty} \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $ by trying to interchange the limits after showing that $\frac{f(x + h) - f(x)}{h}$ converges uniformly as $x \to \infty$. But I couldn't proceed further.
Trying to work backwards, as a specific example, $f(x) = \arctan(x)$ came to my mind. It's derivative certainly goes to $0$ as $x \to \infty$. Doesn't this show that $\frac{f(x + h) - f(x)}{h}$ converges uniformly to $0$ as $x \to \infty$ ?
I'm totally confused! I would really appreciate if anyone told me what I am doing wrong.