Find $\limsup\limits_{x\rightarrow\infty}\ (\sin(x)+\sin(\pi x))$ I'm studying for my Qualifying exams and this was one of the questions in the question bank under real analysis section. I'm currently stuck on this question. I think the answer is 2 but don't have a rigorous proof.
Find $\limsup\limits_{x\rightarrow\infty}\ (\sin(x)+\sin(\pi x))$.
My attempt: I tried to look at the sequence $x_n=\frac{1}{2}+2n$ but not sure how to calculate further and find the $\limsup$
 A: Here is the alternate answer I offered a while ago. The beginning is identical, but from $(3)$ on is an approach not using continued fractions.
Your original idea of looking at $x=\frac12+2n$ is in the right direction, because that gives $\sin(\pi x)=1$. The remaining piece is to seek $x\approx\left(\frac12+m\right)\pi$ so that $\sin(x)\approx1$.

What We Would Like
If $|\,(4m+1)\pi-(4n+1)\,|\le2\epsilon$, then
$$
\begin{align}
1-\sin\left(2n+\tfrac12\right)
&=\left|\,\sin\left(\left(2m+\tfrac12\right)\pi\right)-\sin\left(2n+\tfrac12\right)\tag{1a}\,\right|\\[6pt]
&\le\left|\,\left(2m+\tfrac12\right)\pi-\left(2n+\tfrac12\right)\,\right|\tag{1b}\\[6pt]
&\le\epsilon\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $1-\sin(x)\ge0$ and $\sin\left(\left(2m+\tfrac12\right)\pi\right)=1$
$\text{(1b)}$:  $|\sin(x)-\sin(y)|=2\left|\,\cos\left(\frac{x+y}2\right)\sin\left(\frac{x-y}2\right)\,\right|\le|x-y|$
$\text{(1c)}$: assumption about $m$ and $n$
Since $\sin\left(\left(2n+\tfrac12\right)\pi\right)=1$, we have
$$
\overbrace{\sin\left(2n+\tfrac12\right)}^{\ge1-\epsilon}+\overbrace{\sin\left(\left(2n+\tfrac12\right)\pi\right)}^{=1}\ge2-\epsilon\tag2
$$
Thus, given an $\epsilon\gt0$, we would like to find $m,n\in\mathbb{Z}$ so that $|\,(4m+1)\pi-(4n+1)\,|\le2\epsilon$.

Pigeonhole Approximations
Suppose $\pi$ is a positive irrational number. Choose an $N$ and break $[0,1)$ into $N$ equal sub-intervals
$$
\left\{I_k=\left[\frac kN,\frac{k+1}N\right):0\le k\lt N\right\}\tag3
$$
Using $\{x\}=x-\lfloor x\rfloor$, consider the $N+1$ real numbers
$$
\{\{j\pi\}:0\le j\le N\}\tag4
$$
which live in $[0,1)$. The Pigeonhole Principle says that there must be one of the $N$ sub-intervals, $I_k$, that contains at least two of the $N+1$ real numbers, $\{j_1\pi\}\lt\{j_2\pi\}$. Note that this does not imply that $j_1\lt j_2$, only that
$$
0\lt\{(j_2-j_1)\pi\}\lt\frac1N\tag5
$$
$(5)$ says that there are integers $p,q\in\mathbb{Z}$, with $|q|\le N$, so that
$$
0\lt p+q\pi\lt\frac1N\tag6
$$
The set
$$
\begin{align}
M
&=\left\{j(p+q\pi):1\le j\lt\frac1{p+q\pi}\right\}\tag{7a}\\
&=\left\{\{jq\pi\}:1\le j\lt\frac1{\{q\pi\}}\right\}\tag{7b}
\end{align}
$$ has at least one element in each $I_k$ since the distance between elements of $M$ is $p+q\pi\lt\frac1N$. Furthermore, $M$ covers all of the $I_k$ because $j\gt\frac1{p+q\pi}\implies j(p+q\pi)\gt1$.
Since $N$ can be as large as we wish, we have shown that $\{\mathbb{Z}\pi\}$ is dense in $[0,1]$.

Getting What We Want
We can rewrite the condition $|\,(4m+1)\pi-(4n+1)\,|\le2\epsilon$ from above as
$$
\begin{align}
\frac\epsilon2
&\ge\left|\,m\pi-n+\frac{\pi-1}4\right|\tag{8a}\\
&=\left|\{m\pi\}-\frac{5-\pi}4\right|\tag{8b}
\end{align}
$$
We can find an $m$ to satisfy $(8)$ because $\{\mathbb{Z}\pi\}$ is dense in $[0,1]$.
A: You're off to a good start.  If $x = 2n+\frac12$ then $\sin \pi x=1$.  How big can we make $\sin x$ though?  We'd like to say $x\approx 2m\pi+\frac\pi2$ so that $\sin x\approx 1$ and then conclude that the $\limsup$ is $2$.  Then we'd have $$\pi\approx\frac{2n+1/2}{2m+1/2}=\frac{4n+1}{4m+1}$$  So, that's the plan of attack.  Show that there is an infinite sequence of fractions of this form converging to $\pi$ and then conclude that there is a sequence $(x_n)$ such that $$\lim_{n\to\infty}\sin(x_n)+\sin(\pi x_n)=2$$
Can you carry out the plan?
EDIT
The above is incorrect.  See the correction by Thomas Andrews in the comments.
A: Not an answer, but too,long for comment.
$2$ is certainly an upper bound, and is likely the supremes.  What you want is:
$$x\approx \frac{\pi}{2}(4n+1)\\
\pi x\approx \frac{\pi}{2}(4m  +1)$$
If $x=\frac 12(4m+1)$ then you need:
$$\pi\approx \frac{4m+1}{4n+1}\tag 1$$
The question is, how good an approximation can we get in this form? If $(4m+1)/(4n+1)$ is in the continued fraction for $\pi,$ then:$$-\frac1{2(4n+1)}<\frac12(4n+1)\pi-\frac12 (4m+1)<\frac1{2(4n+1)}$$
In which case $$\sin ((4m+1)/2)>1-\frac1{4(4n+1 )^2}$$
So if there are infinitely many such continued fractions, we can get closer and closer to $2.$
But I can’t think of any way to ensure the continued fractions will take this form infinitely often. Given the apparent randomness of the coefficients for $\pi,$ we might expect a convergent of this form in one out of ever $12$ convergents, but proving that will be hard.
The first example is:$$\frac{208341}{66317}$$ and $x=\frac{208341}{2},$ and, according to my calculator, $1-\sin x <\frac1{10^{11}}.$
The next is $$\frac{165707065}{52746197}.$$
But proving there are infinitely many will be hard.
So we might need less stringent conditions on $m,n.$ We need a sequence $(m_i,n_i)$ such that:
$$\pi(4m_i+1)-(4n_i+1)\to 0.$$
