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.
Best Answer
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.