Limit of $\sin\left(\sqrt {x+2}\right)-\sin\left(\sqrt{x+4}\right)$ as $x\to\infty$ I wish to find the limit of $$\sin\left(\sqrt {x+2}\right)-\sin\left(\sqrt{x+4}\right)$$ as $x\to\infty$. I think that this limit does not exist.
 A: Original question:
\begin{align*}
\sin(\sqrt{x+2}-\sqrt{x+4})=\sin\left(\dfrac{-2}{\sqrt{x+2}+\sqrt{x+4}}\right)\rightarrow\sin(0)=0.
\end{align*}
A: You can rewrite it as follows:
$$\lim_{x \to \infty} \sin{(\frac{x+2-(x+4)}{\sqrt{x+2}+\sqrt{x+4}})}=
\lim_{x \to \infty}\sin{(\frac{-2}{\sqrt{x+2}+\sqrt{x+4}})}=\sin{0}=0$$
A: Since
$$
\sin p - \sin q = 2\cos \left( {\frac{{p + q}}
{2}} \right)\sin \left( {\frac{{p - q}}
{2}} \right)
$$
you have that
$$
\begin{gathered}
  \left| {\sin p - \sin q} \right| = 2\left| {\cos \left( {\frac{{p + q}}
{2}} \right)\sin \left( {\frac{{p - q}}
{2}} \right)} \right| =  \hfill \\
   \hfill \\
   = 2\left| {\sin \left( {\frac{{p - q}}
{2}} \right)\cos \left( {\frac{{p + q}}
{2}} \right)} \right| \leqslant 2\left| {\frac{{p - q}}
{2}} \right| \hfill \\ 
\end{gathered} 
$$
thus
$$
\left| {\sin \left( {\sqrt {x + 2} } \right) - \sin \left( {\sqrt {x + 4} } \right)} \right| \leqslant \left| {\sqrt {x + 2}  - \sqrt {x + 4} } \right|
$$
But
$$
\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {x + 2}  - \sqrt {x + 4} } \right) = 0
$$
so 
$$
\mathop {\lim }\limits_{x \to  + \infty } \left| {\sin \left( {\sqrt {x + 2} } \right) - \sin \left( {\sqrt {x + 4} } \right)} \right| = 0
$$
and
$$
\mathop {\lim }\limits_{x \to  + \infty } \sin \left( {\sqrt {x + 2} } \right) - \sin \left( {\sqrt {x + 4} } \right) = 0
$$
A: In general,
$$|\sin A-\sin B|=\left|\int_A^B\cos t\,dt\right|\le|A-B|.$$
So if $A(x)-B(x)\to0$, then $\sin A(x)-\sin B(x)\to0$. Take $A(x)=\sqrt{x+2}$
and $B(x)=\sqrt{x+4}$ in this example.
A: Hint: $\sin A -\sin B =2\cos (\frac {A+B} 2) \sin (\frac {A-B} 2)$ and hence $|\sin (\sqrt {x+2} -\sin (\sqrt {x+4}) |\leq 2 |\sin (\sqrt {x+2} - \sqrt {x+4})/2| \to 0$ since  $\sqrt {x+2} - \sqrt {x+4} =\frac {-2} {\sqrt {x+2} +\sqrt {x+4}} \to 0$
