<s>[INST] 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.
 [/INST] 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
$$
\mathop {\lim }\limits_{x \to  + \infty } \sin \left( {\sqrt {x + 2} } \right) - \sin \left( {\sqrt {x + 4} } \right) = 0
$$
</s>