1
$\begingroup$

I need help with some limit I got stuck with. $$\lim_{x\to +\infty} \frac{\cos\left(\frac{\sqrt{x+1}+\sqrt{x}}{2}\right)}{\sin\left(\frac{\sqrt{x+1}-\sqrt{x}}{2}\right)} - 4\sqrt{x+1}\cos\left(\sqrt{x+1}\right)-\sin\left(\sqrt{x+1}\right)$$

When I calculate this for large numbers, it seem like it tends toward zero. As an attempt to prove it, I squared the equation in WolframAlpha and after some simplifying: $$\left(\frac{\cos\left(\frac{\sqrt{x+1}+\sqrt{x}}{2}\right)}{\sin\left(\frac{\sqrt{x+1}-\sqrt{x}}{2}\right)} - 4\sqrt{x+1}\cos\left(\sqrt{x+1}\right)-\sin\left(\sqrt{x+1}\right)\right)^2 = \frac{1+\cos\left(2\sqrt{x+1}\right)}{2}\left(4\sqrt{x+1}-\cot\left(\frac{\sqrt{x+1} - \sqrt{x}}{2}\right)\right)^2$$

noticing that: $$0 \leq \frac{1+\cos\left(2\sqrt{x+1}\right)}{2} \leq 1$$

so that means the above limit is related to this new limit: $$\lim_{x\to +\infty} 4\sqrt{x+1}-\cot\left(\frac{\sqrt{x+1} - \sqrt{x}}{2}\right)$$ which also seem to tend toward zero but don't know how to prove it, any help?

$\endgroup$
4
  • 1
    $\begingroup$ What let you suppose that limit exists? $\endgroup$
    – user
    May 11 at 10:12
  • $\begingroup$ it's my conjecture and the secend limit have a direct relation to it. $\endgroup$
    – AnisSaouliBH
    May 11 at 10:16
  • $\begingroup$ so i need who help me with it. $\endgroup$
    – AnisSaouliBH
    May 11 at 10:16
  • $\begingroup$ Could you share your evaluation by WA? $\endgroup$
    – user
    May 11 at 10:24

2 Answers 2

Reset to default
3
$\begingroup$

$$\frac{\sqrt{x+1}-\sqrt{x}}2=u, u\to 0$$ $$a=\sqrt{x+1}=u+\frac{1}{4u}$$ $$\frac{\cos\left(\frac{\sqrt{x+1}+\sqrt{x}}{2}\right)}{\sin\left(\frac{\sqrt{x+1}-\sqrt{x}}{2}\right)} - 4\sqrt{x+1}\cos\left(\sqrt{x+1}\right)-\sin\left(\sqrt{x+1}\right)=$$ $$=\frac{\cos(a-u)}{\sin u}-4a\cos a-\sin a= \frac{\cos a (\cos u- 4a \sin u)}{\sin u}=(\cot u-4a)\cos a;$$ $$\cot u-4a=\cot u-4u-\frac1{u}=O(u)\to 0$$

$\endgroup$
1
  • $\begingroup$ Very nice way to handle it! $\endgroup$
    – user
    May 11 at 10:37
1
$\begingroup$

With reference with the last limit, we have that

$$\frac{\sqrt{x+1}-\sqrt{x}}{2}=\frac{1}{2\left(\sqrt{x+1}+\sqrt{x}\right)}$$

therefore

$$\cot\left(\frac{\sqrt{x+1} - \sqrt{x}}{2}\right)=\frac1{\tan\left(\frac{\sqrt{x+1} - \sqrt{x}}{2}\right)}=\frac1{\frac{1}{2\left(\sqrt{x+1}+\sqrt{x}\right)}+O(x\sqrt x)}=4\sqrt x+O\left(\frac1x\right)$$

and then

$$ 4\sqrt{x+1}-\cot\left(\frac{\sqrt{x+1} - \sqrt{x}}{2}\right)=4\sqrt{x+1}-4\sqrt x+O\left(\frac1x\right) \to 0$$

$\endgroup$
3
  • $\begingroup$ would you please try to exlpain what you did after 1/tan() $\endgroup$
    – AnisSaouliBH
    May 11 at 10:26
  • $\begingroup$ i know what a O() is but i need you to explain the two steps after 1/tan() $\endgroup$
    – AnisSaouliBH
    May 11 at 10:27
  • $\begingroup$ We have used that $\tan t=t+O(t^3)$ for $t\to 0$ $\endgroup$
    – user
    May 11 at 10:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .