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?
 A: $$\left|\sin\sqrt{x+1}-\sin\sqrt{x}\right|=\left|\int_{\sqrt{x}}^{\sqrt{x+1}}\cos u\,du\right|\leq\int_{\sqrt{x}}^{\sqrt{x+1}}1\,du = O\left(\frac{1}{\sqrt{x}}\right).$$
A: Hint: $$\sin(p)-\sin(q)=2\sin(\frac{p-q}{2})\cos(\frac{p+q}{2})$$
A: 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.
A: i think the limit is zero. here is the reason:
$$\begin{align}\sin\sqrt{x+1} - \sin \sqrt x  &= 2\cos((\sqrt{x+1}+\sqrt x)/2)\sin(\sqrt{x+1} - \sqrt x)/2)\\
&=2\cos(\sqrt x + \cdots)\sin(1/(4\sqrt x)+\cdots)\\
&=\frac1{4\sqrt x}\cos(\sqrt x+\cdots) \to 0   \text{  as } x \to \infty.\end{align}$$ 

p.s. in fact, we don't need the trig identity. we use $$\sqrt{x+1} - \sqrt x = \frac 1{2\sqrt x} + \cdots .$$ this tells us that the arc length between the terminal points $(\cos \sqrt{x+1}, \sin \sqrt{x+1})$ and $(\cos x, \sin x)$ on the unit circle goes to zero. therefore the distance between the $y$-coordinates which is smaller than the arc length must also go to zero. 
A: Hint:
Mean value theorem implies there exist $c\in]x,x+1[$ such that
$$\sin(\sqrt{x+1})-\sin(\sqrt{x})=\frac{\sqrt{x+1}-\sqrt{x}}{2\sqrt{c}}\cos\sqrt{c}$$
Then $$\left|\sin(\sqrt{x+1})-\sin(\sqrt{x})\right|\le\left(\frac{\sqrt{x+1}}{2\sqrt{x}}-\frac{\sqrt{x}}{2\sqrt{x+1}}\right)|\cos\sqrt{c}|\le\frac{1}{2}\left(\sqrt{1+\frac{1}{x}}-\sqrt{\frac{1}{1+1/x}}\right)$$
A: Hint: For $x$ positive, we have $\sqrt{x}\lt \sqrt{x+1}\lt \sqrt{x}+\frac{1}{2\sqrt{x}}$.
A: 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 $(\sin \sqrt {x+1}-\sin \sqrt x) \to 0.$
