Evaluate $\lim_{n\to\infty} \sin^2 (\pi\sqrt{n^{2014} + n^{2012} + 1})$ I tried to use $\sin^2(\pi x) = \sin^2 (\pi x - \pi)$
So,
$$\lim_{n\to\infty}\sin^2 (\pi\sqrt{n^{2014} + n^{2012} + 1}) \\
=\lim_{n\to\infty}\sin^2 (\pi(\sqrt{n^{2014} + n^{2012} + 1} - 1)) \\
\lim_{n\to\infty}\sin^2 (\frac{\pi (n^{2014} + n^{2012})}{\sqrt{n^{2014} + n^{2012} + 1}})$$
And I got trouble on here, Thanks for any hint.
 A: NOTE: this is not a complete answer but a small collection of some partial results.
First remark
$$
\begin{gathered}
  \pi \sqrt {n^2  + n + 1}  = \pi n\left( {1 + \frac{{\text{1}}}
{n} + \frac{1}
{{n^2 }}} \right)^{1/2}  =  \hfill \\
   \hfill \\
   = \pi n\left( {1 + \frac{1}
{{2n}} + o\left( {\frac{1}
{n}} \right)} \right) = \pi n + \frac{\pi }
{2} + o\left( 1 \right) \hfill \\ 
\end{gathered} 
$$
this means that
$$
\begin{gathered}
  \sin ^2 \left( {\pi \sqrt {n^2  + n + 1} } \right) = \sin ^2 \left( {\pi n + \frac{\pi }
{2} + o\left( 1 \right)} \right) =  \hfill \\
   \hfill \\
   = \sin ^2 \left( {\frac{\pi }
{2} + o\left( 1 \right)} \right) \hfill \\ 
\end{gathered} 
$$
and therefore
$$
\mathop {\lim }\limits_{n \to  + \infty } \left[ {\sin ^2 \left( {\pi \sqrt {n^2  + n + 1} } \right)} \right] = \mathop {\lim }\limits_{n \to  + \infty } \left[ {\sin ^2 \left( {\frac{\pi }
{2} + o\left( 1 \right)} \right)} \right] = 1
$$
This prove that, in general, it is not enough to consider only the main term of the polynomial under the square root.
Second remark
Let us consider a more general case
$$
\begin{gathered}
  \pi \sqrt {n^{2k}  + n^k  + 1}  = \pi n^k \left( {1 + \frac{{\text{1}}}
{{n^k }} + \frac{1}
{{n^{2k} }}} \right)^{1/2}  =  \hfill \\
   \hfill \\
   = \pi n^k \left( {1 + \frac{1}
{{2n^k }} + o\left( {\frac{1}
{{n^k }}} \right)} \right) = \pi n^k  + \frac{\pi }
{2} + o\left( 1 \right) \hfill \\ 
\end{gathered} 
$$
and as before we have that
$$
\mathop {\lim }\limits_{n \to  + \infty } \left[ {\sin ^2 \left( {\pi \sqrt {n^{2k}  + n^k  + 1} } \right)} \right] = \mathop {\lim }\limits_{n \to  + \infty } \left[ {\sin ^2 \left( {\frac{\pi }
{2} + o\left( 1 \right)} \right)} \right] = 1
$$
On the other side, if we consider $n^{2k}+n^h+1$ with $0 \leq h<k$ we have that
$$
\begin{gathered}
  \pi \sqrt {n^{2k}  + n^h  + 1}  = \pi n^k \left( {1 + \frac{{\text{1}}}
{{n^{2k - h} }} + \frac{1}
{{n^{2k} }}} \right)^{1/2}  =  \hfill \\
   \hfill \\
   = \pi n^k \left( {1 + \frac{{\text{1}}}
{{2n^{2k - h} }} + o\left( {\frac{{\text{1}}}
{{n^{2k - h} }}} \right)} \right) = \pi n^k  + \frac{\pi }
{{2n^{k - h} }} + o\left( {\frac{1}
{{n^{k - h} }}} \right) \hfill \\ 
\end{gathered} 
$$
thus
$$
\mathop {\lim }\limits_{n \to  + \infty } \left[ {\sin ^2 \left( {\pi \sqrt {n^{2k}  + n^k  + 1} } \right)} \right] = \mathop {\lim }\limits_{n \to  + \infty } \left[ {\sin ^2 \left( {\frac{\pi }
{{2n^{k - h} }} + o\left( {\frac{1}
{{n^{k - h} }}} \right)} \right)} \right] = 0
$$
Third remark:
My conjecture about the case $k<h<2k$ is motivated by numerical trials like  that below which is relative to the case $n^{20}+n^{18}+1$

I have some ideas but not yet a proof. Of course, if my conjecture is true the proposed limit does not exists.
Fourth remark:
Let be $h=k+1$ with $k \geq 3$. Then
$$
\begin{gathered}
  \pi \sqrt {n^{2k}  + n^{k + 1}  + 1}  = \pi n^k \left( {1 + \frac{{\text{1}}}
{{n^{k - 1} }} + \frac{1}
{{n^{2k} }}} \right)^{1/2}  =  \hfill \\
   \hfill \\
   = \pi n^k \left( {1 + \frac{{\text{1}}}
{{2n^{k - 1} }} - \frac{1}
{{8n^{2k - 2} }} + o\left( {\frac{1}
{{n^{2k - 2} }}} \right)} \right) =  \hfill \\
   \hfill \\
   = \pi n^k  + \frac{{\pi n}}
{2} - \frac{1}
{{8n^{k - 2} }} + O\left( {\frac{1}
{{n^{k - 2} }}} \right) \hfill \\
   \hfill \\ 
\end{gathered} 
$$
Therefore
$$
\begin{gathered}
  \sin ^2 \left( {\pi \sqrt {n^{2k}  + n^{k + 1}  + 1} } \right) = \sin ^2 \left( {\pi \sqrt {n^{2k}  + n^{k + 1}  + 1} } \right) \hfill \\
   = \sin ^2 \left( {\pi n^k  + \frac{{\pi n}}
{2} - \frac{1}
{{8n^{k - 2} }} + O\left( {\frac{1}
{{n^{k - 2} }}} \right)} \right) =  \hfill \\
   \hfill \\
   = \sin ^2 \left( {\frac{{\pi n}}
{2} - \frac{1}
{{8n^{k - 2} }} + O\left( {\frac{1}
{{n^{k - 2} }}} \right)} \right) \hfill \\
   \hfill \\ 
\end{gathered} 
$$
so that the limit does not exists.
A: I think the following argument is correct:
Imagine the sequence $u_n=\sin (\pi \sqrt{n^2+1}).$
It's easy to understand that $\lim_{n\to \infty}u_n=0$.
You can see it on the next representation of $u_n$:

That's because, when $n \to \infty$, you have $n^2+1 \sim n^2$ and you may write:
$$\lim_{n\to \infty}\sin (\pi \sqrt{n^2+1})=\lim_{n\to \infty}\sin (\pi \sqrt{n^2})=\lim_{n\to \infty}\sin (\pi n)=\lim_{n\to \infty}0=0$$
A similar argument may prove you that $\lim_{n\to\infty} \sin^2 (\pi\sqrt{n^{2014} + n^{2012} + 1})=0$, since $\sqrt{n^{2014} + n^{2012} + 1} \sim n^{1007}$ (when $n \to \infty$) which is always a natural number.
If there is a flaw on my argument, please let me know and I will delete this answer.
EDIT:
Actually, you can evaluate this using just the limits properties:
$$\begin{align*}
\lim_{n\to\infty} \sin^2 (\pi\sqrt{n^{2014} + n^{2012} + 1})&=\sin^2 \left( \lim_{n\to\infty}(\pi\sqrt{n^{2014} + n^{2012} + 1}) \right)\\
&=\sin^2 \left(\pi \lim_{n\to\infty}(\sqrt{n^{2014} + n^{2012} + 1}) \right)\\
&=\sin^2 \left(\pi \sqrt{\lim_{n\to\infty}(n^{2014} + n^{2012} + 1}) \right)\\
&=\sin^2 \left(\pi \sqrt{\lim_{n\to\infty}n^{2014} } \right)\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;(1)\\
&=\sin^2 \left(\pi \lim_{n\to\infty}\sqrt{n^{2014} } \right)\\
&=\sin^2 \left(\pi \lim_{n\to\infty}{n^{1007} } \right)\\
&=\lim_{n\to\infty}\left(\sin^2 \left(\pi {n^{1007} } \right)\right)\\
&=\lim_{n\to\infty}\left(0^2\right)=\lim_{n\to\infty}0=0\\
\end{align*}$$
I used approximation before to justifying a known theorem $(1)$:

If $f(x)=a_0x^n+a_1x^{n-1}+ ... +a_n$ where $n\in \mathbb{N}$ and $a_0\neq0$, then:
  $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}a_0x^n$$

which you use on the simplifications above.
