Evaluate the limit $\;\lim_{n \to \infty} \sin\bigg(\pi\sqrt{n^2+n-3}\bigg)$

Question

Evaluate the limit$$\lim_{n \to \infty} \sin\bigg(\pi\sqrt{n^2+n-3}\bigg)$$ and $n\in N$

My Approach:

$\lim_{n \to \infty} \sin\bigg(\pi\sqrt{n^2+n-3}\bigg)\implies \lim_{n \to \infty} {\pm}\sin\bigg(n\pi-\pi\sqrt{n^2+n-3}\bigg)$

$\implies \pm\lim_{n \to \infty} \sin\bigg[\pi\bigg(n-\sqrt{n^2+n-3}\bigg)\bigg]$

$\pm\lim_{n \to \infty} \sin\left[\dfrac{\pi\bigg(n-\sqrt{n^2+n-3}\bigg)\bigg(n+\sqrt{n^2+n-3}\bigg)}{(n+\sqrt{n^2+n-3}\bigg)}\right]$

$\implies \pm\lim_{n \to \infty} \sin\left[\dfrac{\pi\bigg(n^{2}-\left(\sqrt{n^2+n-3}\right)^2\bigg)}{(n+\sqrt{n^2+n-3}\bigg)}\right]$

$\implies \pm \lim_{n\to \infty}\sin\left[\dfrac{\pi\left(-n+3\right)}{n\left(1+\sqrt{1+\dfrac{1}{n}-\dfrac{3}{n^{2}}}\right)}\right]$

$\implies \pm \lim_{n\to \infty} \sin\left[\dfrac{\pi n \left(-1+\dfrac{3}{n}\right)}{n\left(1+\sqrt{1+\dfrac{1}{n}-\dfrac{3}{n^{2}}}\right)}\right]$

$\implies \pm \sin\left(\dfrac{-\pi}{2}\right)\implies\pm 1$

Doubt: Is my solution and answer correct because my book does not contain answer to this question

Other Approaches are also welcomed.

Answer 1

We have $\sqrt{n^2 + n - 3} = n\Big(1 + \frac{n-3}{n^2}\Big)^{1/2}$

By binomial expansion, we have

$$ (1 + x)^{\frac{1}{2}} = 1 + \frac{x}{2} - \frac{x^2}{8} + \ldots $$

Plug in $x = \frac{n-3}{n^2}$ and calculate the first two terms and simplify to get

$$ \sqrt{n^2 + n - 3} = n + \frac{1}{2} + \frac{13}{8n} + \ldots $$

When $n \to \infty$ every term after the first two terms will vanish. Hence for $n \in N$,

$$ \lim_{n \to \infty}\sin(\pi\sqrt{n^2 + n - 3}) = \lim_{n \to \infty}\sin(n\pi + \pi/2)) = \lim_{n \to \infty}(-1)^n $$

Hence the limit does not exist as it will either be $+1$ or $-1$ depending of if $n$ is even or odd.

Answer 2

(A) $L=\lim_{n \to \infty} \sin\bigg(\pi\sqrt{n^2+n-3}\bigg)$ where $n$ is real : Does not Exist.

(B) $L=\lim_{n \to \infty} \sin\bigg(\pi\sqrt{n^2+n-3}\bigg)$ where $n$ is Even Natural : $L=+\sin(\pi/2)=+1$.

We can see that $L=\lim_{n \to \infty} \sin\bigg(\pi\sqrt{n^2+n-3}\bigg)$ gives :

$L=\lim_{n \to \infty} \sin\bigg(\pi\sqrt{n^2+n+1/4-1/4-3}\bigg)$

$L=\lim_{n \to \infty} \sin\bigg(\pi\sqrt{(n+1/2)^2-13/4}\bigg)$

$L=\lim_{n \to \infty} \sin\bigg(\pi(n+1/2)\bigg)$

$L=\lim_{n \to \infty} \sin\bigg(\pi n + \pi / 2\bigg)$

$L=\pm \sin(\pi / 2)$ ( We can think that in the limit , we get a sequence where the value toggles between $+1$ & $-1$ , though strictly speaking , there is no Unique Limit in this Case here )

A little tweaking in the Exercise will give the Unique Limit :

We can make it that $n$ is Even Natural , then
$L=+\sin(\pi / 2)=+1$

Answer 3

We have that

$$ \pi n-\pi\sqrt{n^2+n-3} =\pi \frac{3-n}{n+\sqrt{n^2+n-3}} \to-\frac \pi 2$$

then for $n$ odd

$$\sin\left(\pi\sqrt{n^2+n-3}\right)=\sin\left(\pi n-\pi\sqrt{n^2+n-3}\right) \to -1$$

while for $n$ even

$$\sin\left(\pi\sqrt{n^2+n-3}\right)=-\sin\left(\pi n-\pi\sqrt{n^2+n-3}\right) \to 1$$