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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.

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    $\begingroup$ You have created a lot of confusion by edited and re-editing the question, $\endgroup$ – Kavi Rama Murthy Dec 20 '19 at 5:51
  • $\begingroup$ Sorry for re-editing. $\endgroup$ – N math Dec 20 '19 at 5:56
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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*}

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You can rewrite it as follows:

$$\lim_{x \to \infty} \sin{(\frac{x+2-(x+4)}{\sqrt{x+2}+\sqrt{x+4}})}= \lim_{x \to \infty}\sin{(\frac{-2}{\sqrt{x+2}+\sqrt{x+4}})}=\sin{0}=0$$

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  • $\begingroup$ I edited the question. $\endgroup$ – N math Dec 20 '19 at 5:40
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    $\begingroup$ You already got two good answers, now it's a different question $\endgroup$ – Andronicus Dec 20 '19 at 5:40
  • $\begingroup$ @Nmath you can post another one $\endgroup$ – Andronicus Dec 20 '19 at 5:41
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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 $$ \mathop {\lim }\limits_{x \to + \infty } \sin \left( {\sqrt {x + 2} } \right) - \sin \left( {\sqrt {x + 4} } \right) = 0 $$

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  • $\begingroup$ Great. Thanks for your answer. $\endgroup$ – N math Dec 20 '19 at 5:53
  • $\begingroup$ You are welcome! $\endgroup$ – Luca Goldoni Ph.D. Dec 20 '19 at 5:55
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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.

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  • $\begingroup$ Great. Thanks for your answer. $\endgroup$ – N math Dec 20 '19 at 7:22
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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 $\sqrt {x+2} - \sqrt {x+4} =\frac {-2} {\sqrt {x+2} +\sqrt {x+4}} \to 0$

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  • $\begingroup$ Thanks for your answer. $\endgroup$ – N math Dec 20 '19 at 5:54

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