# What is lim_(nrarroo)(-1)^(n-1)sin(pisqrt(n^2+0.5n+1)) ?

Mar 25, 2016

The limit is $- \frac{1}{\sqrt{2}}$

#### Explanation:

$\text{ }$

We need to find the asymptote to $\sqrt{{n}^{2} + 0.5 n + 1}$.
(Thanks, George C.)

To find (linear) oblique asymptote $y = m n + b$,

$m = {\lim}_{n \rightarrow \infty} \frac{\sqrt{{n}^{2} + 0.5 n + 1}}{n} = 1$.

Now we find $b$:

$b = {\lim}_{n \rightarrow \infty} \frac{\sqrt{{n}^{2} + 0.5 n + 1}}{m} n = {\lim}_{n \rightarrow \infty} \frac{\sqrt{{n}^{2} + 0.5 n + 1}}{n} = \frac{1}{4}$.

So, $\sqrt{{n}^{2} + 0.5 n + 1}$ is asymptotic to $n + \frac{1}{4}$.

Alternative method from George C

$\sqrt{{n}^{2} + 0.5 n + 1} = \sqrt{{\left(n + \frac{1}{4}\right)}^{2} + \frac{15}{16}}$

 = (n+1/4)sqrt(1+15/(16(n+1/4)^2)

(If I've deduced the method correctly, our goal is to get $\left(m n + b\right) \sqrt{1 + g \left(n\right)}$ where ${\lim}_{n \rightarrow \infty} g \left(n\right) = 0$.)

Returning to the big question

${\lim}_{n \rightarrow \infty} {\left(- 1\right)}^{n - 1} \sin \left(\pi \sqrt{{n}^{2} + 0.5 n + 1}\right) = {\lim}_{n \rightarrow \infty} {\left(- 1\right)}^{n - 1} \sin \left(\pi \left(n + \frac{1}{4}\right) \sqrt{1 + \frac{15}{16 {\left(n + \frac{1}{4}\right)}^{2}}}\right)$

For even $n$, say $n = 2 k$,
we get $- \sin \left(2 k \pi + \frac{\pi}{4}\right) = - \sin \left(\frac{\pi}{4}\right) = - \frac{1}{\sqrt{2}}$

For odd $n$, say $n = 2 k + 1$,
we get $\sin \left(2 k \pi + \pi + \frac{\pi}{4}\right) = \sin \left(\frac{5 \pi}{4}\right) = - \frac{1}{\sqrt{2}}$

Therefore,

${\lim}_{n \rightarrow \infty} {\left(- 1\right)}^{n - 1} \sin \left(\pi \sqrt{{n}^{2} + 0.5 n + 1}\right) = - \frac{1}{\sqrt{2}}$

Mar 25, 2016

$- \frac{\sqrt{2}}{2}$

#### Explanation:

Extra care should be taken since we are dealing with a periodic function of the kind
$f \left(x\right) = \pm \sin x$

It's not sufficient that, when $n \to \infty$, ${n}^{2}$ is much greater than $0.5 n + 1$, because if $\sqrt{{n}^{2} + 0.5 n + 1} - n > 0$ than the function will have a value possibly different than zero.

Intuitively it's easy to see that $\sqrt{k + 1} \to \sqrt{k}$ (or, even more importantly for the case, that $\sqrt{k + 1} - \sqrt{k} \to 0$) when $k \to \infty$
Or
$\sqrt{1001} - \sqrt{1000} = 0.015 , 807$
and
$\sqrt{1 , 000 , 001} - \sqrt{1 , 000 , 000} = 0.000 , 500$
and so on.

But let's prove that $\sqrt{{n}^{2} + 0.5 n + a} \to \sqrt{{n}^{2} + 0.5 n}$ when $n \to \infty$, where $n \in \mathbb{N}$ and $a \in \mathbb{R}$
${\lim}_{n \to \infty} \left(\sqrt{{n}^{2} + 0.5 n + a} - \sqrt{{n}^{2} + 0.5 n}\right) \cdot \frac{\left(\sqrt{{n}^{2} + 0.5 n + a} + \sqrt{{n}^{2} + 0.5 n}\right)}{\sqrt{{n}^{2} + 0.5 n + a} + \sqrt{{n}^{2} + 0.5 n}} =$
$= {\lim}_{n \to \infty} \frac{\cancel{{n}^{2}} + \cancel{0.5 n} + a - \cancel{{n}^{2}} - \cancel{0.5 n}}{\sqrt{{n}^{2} + 0.5 n + a} + \sqrt{{n}^{2} + 0.5 n}}$
$= {\lim}_{n \to \infty} \frac{a}{\sqrt{{n}^{2} + 0.5 n + a} + \sqrt{{n}^{2} + 0.5 n}}$
$= {\lim}_{n \to \infty} \frac{a}{\infty} = 0$

Since the value of $a$ (in the present case $a = 1$) doesn't matter to the result, we can make
${\lim}_{n \to \infty} \sqrt{{n}^{2} + 0.5 n + 1} = {\lim}_{n \to \infty} \sqrt{{n}^{2} + 0.5 n + \frac{1}{16}} = {\lim}_{n \to \infty} \sqrt{{\left(n + \frac{1}{4}\right)}^{2}} = {\lim}_{n \to \infty} \left(n + \frac{1}{4}\right)$

So the main expression becomes

$= {\lim}_{n \to \infty} {\left(- 1\right)}^{n - 1} \sin \left[\pi \left(n + \frac{1}{4}\right)\right]$
$= {\lim}_{n \to \infty} {\left(- 1\right)}^{n - 1} \sin \left(n \cdot \pi + \frac{\pi}{4}\right)$
But
$\sin \left(n \cdot \pi + \frac{\pi}{4}\right) = \sin \left(n \cdot \pi\right) \cdot \cos \left(\frac{\pi}{4}\right) + \sin \left(\frac{\pi}{4}\right) \cdot \cos \left(n . \pi\right) =$
(remembering that $n \in \mathbb{N}$ => $\sin \left(n \cdot \pi\right) = 0$)
$= \frac{\sqrt{2}}{2} \cdot \cos \left(n \cdot \pi\right)$
In the main expression
$= {\lim}_{n \to \infty} {\left(- 1\right)}^{n - 1} \cdot \frac{\sqrt{2}}{2} \cdot \cos \left(n \cdot \pi\right)$

Now consider the possible values of the expression above
If n is odd => (-1)^("even number").sqrt2/2*cos[("odd number")*pi]=1*sqrt2/2(-1)=-sqrt2/2
If n is even => (-1)^("odd number").sqrt2/2*cos[("even number")*pi]=(-1)*sqrt2/2*1=-sqrt2/2

So, for $n \in \mathbb{N}$
$= {\lim}_{n \to \infty} {\left(- 1\right)}^{n - 1} \cdot \frac{\sqrt{2}}{2} \cdot \cos \left(n \cdot \pi\right) = - \frac{\sqrt{2}}{2}$

Mar 27, 2016

$- \frac{1}{\sqrt{2}}$.

#### Explanation:

I have revised my answer, thanks to George for his affirmation on his answer. I am sorry for having overlooked some nicety in applying the notion of 'one is asymptotic with another', in this limit problem.

The correction follows, without changing my approach.

pisqrt(n^2+0.5n+1)=npisqrt(1+0.5/n+1/n^2))=npi(1+(1/2)(0.5/n)+O(1/n^2)), using expansion in powers of $\frac{1}{n} .$
Here, $n \to \infty$ through integer values 1,2,3,..,

The limit is same as the limit for
${\left(- 1\right)}^{n} \sin \left(n \pi + \frac{\pi}{4} + O \left(\frac{1}{n}\right)\right) = - \frac{1}{\sqrt{2}}$, as given in the other answers..