Evaluate $\lim\limits_{x\to\infty}(\sin\sqrt{x+1}-\sin\sqrt{x})$ When using Maclaurin series, the limit is
$$\lim\limits_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}=0$$
If we expand the expression with two limits
$$\lim\limits_{x\to\infty}\sin\sqrt{x+1}-\lim\limits_{x\to\infty}\sin\sqrt{x}$$
it diverges.
Which solution is right?

If $f$ is differentiable on some $[a,\infty)$ and $\lim_{x\to\infty}f'(x) = 0,$ then $\lim_{x\to\infty}(f(x+1) -f(x))= 0.$ Proof: By the MVT, $f(x+1) -f(x) = f'(c_x)\cdot 1.$ As $x\to \infty, c_x \to \infty,$ hence $f'(c_x) \to 0$ and we have the result. Example: If $f(x) = \sin \sqrt x\,$ then $f'(x) = (\cos \sqrt x)/(2\sqrt x) \to 0,$ therefore $(\sin \sqrt {x+1}-\sin \sqrt x) \to 0.$
