5
$\begingroup$

Evaluate the limit

$$\lim_{n\rightarrow \infty}4\sqrt{n}\sin (\pi\sqrt{4n^2+\sqrt{n}})$$

We know that :

$$\sin (\pi-x)=\sin x$$

So we have :

$$\lim_{n\rightarrow \infty}4\sqrt{n}\sin (\pi-\pi\sqrt{4n^2+\sqrt{n}})$$

Now :

$$(\pi-\pi\sqrt{4n^2+\sqrt{n}})\cdot \frac{\pi+\pi\sqrt{4n^2+\sqrt{n}}}{\pi+\pi\sqrt{4n^2+\sqrt{n}}}=\frac{\pi^2-\pi^2(4n^2+\sqrt{n})}{\pi+\pi\sqrt{4n^2+\sqrt{n}}}$$

$$\lim_{n\rightarrow \infty}4\sqrt{n}\sin \left(\frac{\pi^2-\pi^2(4n^2+\sqrt{n})}{\pi+\pi\sqrt{4n^2+\sqrt{n}}}\right)$$

Now what ?

$\endgroup$

2 Answers 2

Reset to default
5
$\begingroup$

I think the following is better. $$\lim_{n\rightarrow+\infty}4\sqrt{n}\sin (\pi\sqrt{4n^2+\sqrt{n}})=-\lim_{n\rightarrow+\infty}4\sqrt{n}\sin \left(2\pi n-\pi\sqrt{4n^2+\sqrt{n}}\right)=$$ $$=\lim_{n\rightarrow+\infty}4\sqrt{n}\sin\frac{\pi\sqrt{n}}{2 n+\sqrt{4n^2+\sqrt{n}}}=\lim_{n\rightarrow+\infty}4\sqrt{n}\sin\frac{\pi}{\sqrt{n}\left(2 +\sqrt{4+\frac{1}{n\sqrt{n}}}\right)}=$$ $$=\lim_{n\rightarrow+\infty}\left(\frac{\sin\frac{\pi}{\sqrt{n}\left(2 +\sqrt{4+\frac{1}{n\sqrt{n}}}\right)}}{\frac{\pi}{\sqrt{n}\left(2 +\sqrt{4+\frac{1}{n\sqrt{n}}}\right)}}\cdot\frac{4\pi}{2 +\sqrt{4+\frac{1}{n\sqrt{n}}}}\right)=\pi.$$

$\endgroup$
3
  • $\begingroup$ Or from the second line $$=\lim_{n\to\infty} 4\sqrt{n}\sin\left(\frac{\pi}{4\sqrt{n}}\right) = \lim_{{\pi\over 4\sqrt{n}}\to 0} \pi \frac{\sin \left( {\pi\over 4\sqrt{n}} \right)}{{\pi\over 4\sqrt{n}}} = \pi$$ $\endgroup$
    – Dylan
    Commented Oct 28, 2017 at 22:03
  • $\begingroup$ @Dylan, what identity did you use? $\endgroup$
    – venrey
    Commented Oct 29, 2017 at 3:50
  • $\begingroup$ $\lim_{x\to 0} \frac{\sin x}{x} = 1$ $\endgroup$
    – Dylan
    Commented Oct 29, 2017 at 4:52
1
$\begingroup$

What you could also do is, for large $n$ by Taylor or binomial expansion $$\sqrt{4n^2+\sqrt{n}}=2n+\frac1 {4 \sqrt n}+O\left( \frac1 {n^2}\right)$$ making $$\sin (\pi\sqrt{4n^2+\sqrt{n}})\sim\sin\left( \frac \pi {4 \sqrt n}\right)\sim \frac \pi {4 \sqrt n}$$

Edit

Just added for your curiosity.

If we push the expansion, we can get "good" estimations for small values of $n$.

Using now $$\sqrt{4n^2+\sqrt{n}}=2n+\frac1 {4 \sqrt n}-\frac{1}{64 n^2}+O\left( \frac1 {n^{7/2}}\right)$$ we should get $$4\sqrt n\,\sin (\pi\sqrt{4n^2+\sqrt{n}})=\pi -\frac {\pi^3}{96n} +O\left( \frac1 {n^{3/2}}\right)$$ and then the following values for small $n$ $$\left( \begin{array}{ccc} n & \text{exact} & \text{approximation} & \Delta \\ 1 & 2.70196 & 2.81861 & -0.116649 \\ 2 & 2.92585 & 2.98010 & -0.054248 \\ 3 & 3.00182 & 3.03393 & -0.032117 \\ 4 & 3.03913 & 3.06085 & -0.021722 \\ 5 & 3.06108 & 3.07700 & -0.015917 \\ 6 & 3.07546 & 3.08776 & -0.012299 \\ 7 & 3.08558 & 3.09545 & -0.009869 \\ 8 & 3.09308 & 3.10122 & -0.008145 \\ 9 & 3.09884 & 3.10571 & -0.006870 \\ 10 & 3.10340 & 3.10929 & -0.005896 \end{array} \right)$$

$\endgroup$

You must log in to answer this question.