# 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.

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

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*}

You can rewrite it as follows:

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

• I edited the question. – N math Dec 20 '19 at 5:40
• You already got two good answers, now it's a different question – Andronicus Dec 20 '19 at 5:40
• @Nmath you can post another one – Andronicus Dec 20 '19 at 5:41

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 $$\mathop {\lim }\limits_{x \to + \infty } \sin \left( {\sqrt {x + 2} } \right) - \sin \left( {\sqrt {x + 4} } \right) = 0$$

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

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.

• Great. Thanks for your answer. – N math Dec 20 '19 at 7:22

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 $$\sqrt {x+2} - \sqrt {x+4} =\frac {-2} {\sqrt {x+2} +\sqrt {x+4}} \to 0$$

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