Derivative goes to zero at infinity, does the function have any special properties? What about the limit of the function at infinity? I was faced with a question asking to evaluate $\lim\limits_{x\to\infty}\sin\sqrt{x+2}-\sin\sqrt{x}$. I see that this function's derivative goes to zero as x goes to infinity. Does this have anything to do with the limit of the function at infinity? Does it correspond to some special property?
Note: I know how to solve the problem. I only mentioned the problem because it made me think of this question.
 A: Let $f(x) = \sin(\sqrt{x+2}) - \sin(\sqrt{x})$. If $f'(x)$ approached any value besides 0 as $x \to \infty$, then you would instantly know $\lim_{x \to \infty} f(x)$ must not exist. Intuitively, if we imagine the function eventually has derivative $\approx 3$, then it's sloping upward and we'll definitely have $f(x) \to +\infty$ as $x \to \infty$.
However, some other intuitive-sounding claims turn out to be false:

*

*If $\lim_{x \to \infty} f'(x) = 0$, that does not necessarily mean $\lim_{x \to \infty} f(x)$ exists. Think about the example $f(x) = \log(x)$.

*If $\lim_{x \to \infty} f'(x)$ does not exist, that does not necessarily mean $\lim_{x \to \infty} f(x)$ does not exist. Think about $f(x) = \frac{\sin(x^2)}{x}$, so that $f'(x) = 2 \cos(x^2) - \frac{\sin(x^2)}{x^2}$.

A: If you want to compute the limit in the question, I would advise using the following property:
$$\sin(A) - \sin(B) = 2\sin\left(\frac{A - B}{2}\right)\cos\left(\frac{A + B}{2}\right).$$
So, your limit becomes
$$2\sin\left(\frac{\sqrt{x + 2} - \sqrt{x}}{2}\right)\cos\left(\frac{\sqrt{x + 2} + \sqrt{x}}{2}\right),$$
or, rationalising the numerator,
$$2\sin\left(\frac{1}{\sqrt{x + 2} + \sqrt{x}}\right)\cos\left(\frac{\sqrt{x + 2} + \sqrt{x}}{2}\right).$$
As $x \to \infty$, the $\sin$ term tends to $0$, while the $\cos$ term remains bounded. As such, the limit is $0$.
