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?

Using the prosthaphaeresis formulae,
$$ \sin{A}-\sin{B} = 2\sin{\tfrac{A+B}{2}} \cos{\tfrac{A-B}{2}}, $$
which gives you
$$ 2\sin{\left( \frac{\sqrt{x+1}-\sqrt{x}}{2} \right)} \cos{\left( \frac{\sqrt{x+1}+\sqrt{x}}{2} \right)} $$
Then you have
$$ (\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x}) = 1+x-x=1, $$
so we have
$$ 2\sin{\left( \frac{1}{2(\sqrt{x+1}+\sqrt{x})} \right)} \cos{\left( \frac{\sqrt{x+1}+\sqrt{x}}{2} \right)} $$
Then the limit is the same as
$$ \lim_{y \to \infty} 2(\cos{\tfrac{1}{2}y} )\sin{\left(\frac{1}{2y}\right)}, $$
and the bracket is bounded, the last term tends to zero since $\sin{z} \to 0$ as $z \to 0$. Hence the whole expression tends to zero.
