A subtle limit question: $\lim _{n \to \infty}\sin(\pi\sqrt {n^2+0.5n+1})$

Evaluate

$$ \lim _{n \to \infty}\sin(\pi\sqrt {n^2+0.5n+1}), $$ where $n\in \mathbf{N}$. This is how I solved: $$ \begin{align} &\lim _{n \to \infty}\sin(n\pi\sqrt {1+0.5/n+1/n^2})\\ &= \sin(n\pi)\,\,\,\,\,\,\,\small\color{grey}[{since \sqrt {1+0.5/n+1/n^2} = 1 , as\,\,n\to\infty]}\\ &=0 \end{align} $$ But the actual answer is $\frac{1}{\sqrt2}$ or $\frac{-1}{\sqrt2}$ depending on whether n is even or odd .

For large $n$,

$$\sqrt{n^2+\frac n2+1}=n\sqrt{1+\frac1{2n}+\frac1{n^2}}\approx n\left(1+\frac1{4n}\right)=n+\frac14.$$ to the first order.

Then the expression tends to $\pm\dfrac1{\sqrt2}$ depending on the parity of $n$ so that the limit cannot exist.

Doing almost what you did $$\sqrt{n^2+\frac{n}{2}+1}=n \sqrt{1+\frac{1}{2 n}+\frac{1}{n^2}}$$ Now, using Taylor expansion or generalized binomial theorem $$\sqrt{1+\frac{1}{2 n}+\frac{1}{n^2}}=1+\frac{1}{4 n}+O\left(\frac{1}{n^2}\right)$$ which makes $$\sin \left(\pi \sqrt{n^2+\frac{n}{2}+1}\right)\sim \sin(n\pi+\frac \pi 4)$$ from which you can conclude.

The situation is more delicate than you think. For example, $1+1/\sqrt n \to 1,$ yet $\sin(n\pi((1+1/\sqrt n))$ has no limit. In fact this sequence is dense in $[-1,1].$

Hint: $\sqrt {1+u} = 1 + u/2 + o(u)$ as $u\to 0.$