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fine the limit:

$$\lim_{ n \to \infty }\sin ^2(\pi\sqrt{n^2+n})=?$$

my try :

$$\lim_{ n \to \infty }\sin ^2(\pi\sqrt{n^2+n})=\sin ^2(\lim_{ n \to \infty }\pi\sqrt{n^2+n})\\\sin^2(\infty)=$$Does not exist

is it right ?

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    $\begingroup$ For large $n$ we have $\sqrt{n^2+n}\approx n+\dfrac12$. And $\sin^2(\pi[n+\dfrac12])\approx \sin^2(\pi/2)=1$. More precisely, use $$\lim_{n\to\infty}(\sqrt{n^2+n}-n)=\frac12,$$ and periodicity of sine squared. $\endgroup$ – Jyrki Lahtonen Feb 23 '17 at 13:51
  • $\begingroup$ So the same trick that was used here (and with other questions on our site). I think this question is a duplicate, but I have promised not to vote to close it as such because my vote is immediately binding. $\endgroup$ – Jyrki Lahtonen Feb 23 '17 at 13:53
  • $\begingroup$ Is $ n \in \mathbb{N}$? $\endgroup$ – Alex Silva Feb 23 '17 at 13:56
  • $\begingroup$ it seems that you are right, but this is not a proof $\endgroup$ – Dr. Sonnhard Graubner Feb 23 '17 at 13:59
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$$\begin{align}\lim_{n\rightarrow \infty}\sin^2\left(\pi\sqrt{n^2+n}\right) &=\lim_{n\rightarrow \infty}\sin^2\left(n\pi-\pi\sqrt{n^2+n}\right)\\ &=\lim_{n\rightarrow \infty}\sin^2\bigg(\frac{-\pi^2 n}{n\pi+\pi\sqrt{n^2+n}}\bigg)\end{align}$$

So $$\lim_{n\rightarrow \infty}\sin^2\left(\pi\sqrt{n^2+n}\right) = 1$$

Because:$$\displaystyle \star \; \lim_{n\rightarrow \infty}\frac{\pi^2 n}{n\pi+\pi\sqrt{n^2+n}} = -\lim_{n\rightarrow \infty}\frac{\pi}{1+\sqrt{1+\frac{1}{n}}} = -\frac{\pi}{2}$$

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  • $\begingroup$ How does the first equality follow? $\endgroup$ – Mussé Redi Feb 23 '17 at 13:43
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    $\begingroup$ Sir, see the last step once again and you could add some helpful information for the OP to understand the conclusion. $\endgroup$ – Rohan Feb 23 '17 at 13:44
  • $\begingroup$ because $\sin ^2(n\pi-x) = \sin^2 x,n\in \mathbb{Z}$ $\endgroup$ – DXT Feb 23 '17 at 13:44
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    $\begingroup$ Where the OP says that $n \in \mathbb{Z}$? $\endgroup$ – Alex Silva Feb 23 '17 at 13:57
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    $\begingroup$ @Alex Silva. $n\in\mathbb{N}$ is obviously implicitly assumed. This is a widely shared convention that the letters $n,m,p,q$ (amongst others) should reprensent integers unless otherwise explicitely stated. $\endgroup$ – G. Fougeron Feb 23 '17 at 14:23
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we have $$\left(\sin(\pi\sqrt{n^2+n}-\pi n+\pi n\right)^2=\left(\sin\left(\frac{\pi}{\sqrt{1+\frac{1}{n}}+1}\right)\cos(\pi n)\right)^2$$ and the searched limit is $1$ if $n$ tends to infinity

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  • $\begingroup$ why ?$\left(\sin(\pi(\sqrt{n^2+n}-\pi n+\pi n\right)^2=\left(\sin\left(\frac{\pi}{\sqrt{1+\frac{1}{n}}+1}\right)\cos(\pi n)\right)^2$ $\endgroup$ – Almot1960 Feb 23 '17 at 14:45
  • $\begingroup$ you must youse the addition formulas $$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$$ and $$\sin(\pi n)=0$$ if $n$ is a whole number $\endgroup$ – Dr. Sonnhard Graubner Feb 23 '17 at 14:53

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