limit of $\lim_{n\to \infty} (-1)^{n-1} \sin(\pi \sqrt{n^2 + 0.5n + 1})$ what am I doing wrong? I have to evaluate $$\lim_{n\to \infty} (-1)^{n-1} \sin(\pi \sqrt{n^2 + 0.5n + 1})\,\,where \,\,n\,\epsilon\,N$$
what I would do is just factor out $n^2$ in the sqrare root, then just make the $\frac{0.5}{n}$ and $\frac{1}{n^2}$ term zero.
$$\sin{\left(\pi\sqrt{n^2(1+\frac{0.5}{n}+\frac{1}{n^2})}\,\right)}$$
n goes to infinity so this becomes :-
$$\lim_{n\to\infty} (-1)^{n-1} \sin{\left(n\pi\right)}$$
now $\sin n\pi$ is just $0$ so the answer should be $0$
But the answer is $\sin{\left(-\frac{\pi}{4}\right)}$
Could someone please explain to me what I am doing wrong
 A: The problem is that as $n\to\infty$, $\sqrt{n^2 +0.5n + 1}$ diverges, and $\sqrt{n^2 +0.5n + 1} -n \neq 0$. We have
\begin{align*}
\lim_{n\to\infty} \sqrt{n^2 +0.5n + 1} -n &= \lim_{n\to\infty} \sqrt{(n + 0.25)^2 + 1 - 0.25^2} - n\\
&= \lim_{n\to\infty} \frac{(\sqrt{n^2 +0.5n + 1} - n)(\sqrt{n^2 +0.5n + 1} + n)}{\sqrt{n^2 +0.5n + 1} + n}\\
&= \lim_{n\to\infty} \frac{0.5n + 1}{\sqrt{n^2 +0.5n + 1} + n} = \frac{1}{4}
\end{align*}
A: hint
Begin by noting that
$$(-1)^{n-1}\sin(n\pi +X)=-\sin(X)$$
then, use the fact that when $ x\to 0$,
$$\sqrt{1+x}=1+\frac x2(1+\epsilon(x))$$
with $\lim_{x\to 0}\epsilon(x)=0$.
So
$$\sqrt{1+\frac{1}{2n}+\frac{1}{n^2}}=1+\frac{1}{4n}(1+\epsilon(n))$$
with $\lim_{n\to+\infty}\epsilon(n)=0$.
A: Before $1+\frac{0.5}{n}+\frac1{n^2}$ can "become" $1$, it is greater than $1$. And so when it multiplies by $n^2$, you get something greater than $n^2$. And it turns out that it is still greater enough than $n^2$ to make a difference.
The hint in the comments is to use $n^2+0.5n+1=(n+0.25)^2+\frac{15}{16}$. Then you have $(n+0.25)\sqrt{1+\frac{15}{16(n+0.25)^2}}$. At this point your intuition works out to be correct, and $1+\frac{15}{16(n+0.25)^2}$ goes to $1$ quickly enough. But more should be done to prove this.
