Find the upper and lower limits of $xf(x)$, as $x\rightarrow \infty$ Define $$f(x)=\int_{x}^{x+1}\sin(t^2)dt$$
Find the upper and lower limits $xf(x)$, as $x\rightarrow \infty$.  
I find the answer as $+1, -1$ since $|\sin(x)| \le 1$. (Of course I calculated that function)
Is that right or did I miss something?
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I solved this way.
$2xf(x)=\cos(x^2)-\cos[(x+1)^2]+r(x)$  where $r(x)=\frac{\cos(x+1)^2}{x+1}-2x\int_{x^2}^{(x+1)^2}\frac{cos(u)}{4u^{3/2}}du$
Therefore $xf(x)=\frac{1}{2}{\cos(x^2)-\cos(x+1)^2}+\frac{r(x)}{2}$
Using trigonomeric formula: $2\sin(a)\sin(b)=\cos(a-b)-\cos(a+b)$
Rewrite $xf(x)=±\sin(x^2+x+\frac{1}{2})\sin(x+\frac{1}{2})+\frac{r(x)}{2}$.
As $x\rightarrow \infty, r(x) \rightarrow 0$.  
Suppose $x^2=2k\pi$ for integer $k$.
To achieve $±1$, we have to show that $x+\frac{1}{2} \rightarrow 2n\pi+\frac{\pi}{2}$ for some $n$ as $x\rightarrow \infty$.
For each $n$, there exists $k$ such that $\sqrt{2\pi k}+\frac{1}{2} < 2n\pi+\frac{\pi}{2} <\sqrt{2\pi (k+1)} +\frac{1}{2} $
Distance between $\sqrt{2\pi k}+\frac{1}{2}$ and $2n\pi+\frac{\pi}{2}$ is at most $\sqrt{2\pi (k+1)} +\frac{1}{2} -\sqrt{2\pi k}+\frac{1}{2}$.
As $k \rightarrow \infty$ the distance becomes arbitrary small.
Therefore $ x \rightarrow \infty$, $xf(x)=±\sin(2n\pi+\frac{\pi}{2})=±1$.

$$
\begin{align}
f(x)
&=\int_x^{x+1}\sin(t^2)\,\mathrm{d}t\\
&=-\int_x^{x+1}\frac1{2t}\,\mathrm{d}\cos(t^2)\\
&=\frac{\cos(x^2)}{2x}-\frac{\cos((x+1)^2)}{2(x+1)}
-\int_x^{x+1}\cos(t^2)\frac1{2t^2}\,\mathrm{d}t\\
&=\frac{\cos(x^2)}{2x}-\frac{\cos((x+1)^2)}{2(x+1)}+O\left(\frac1{x^2}\right)\tag{1}
\end{align}
$$
So
$$
\begin{align}
xf(x)
&=\frac{\cos(x^2)}{2}-\frac{\cos((x+1)^2)}{2(1+1/x)}+O\left(\frac1{x}\right)\\
&=\frac12\left(\cos(x^2)-\cos((x+1)^2)\right)+O\left(\frac1{x}\right)\tag{2}
\end{align}
$$
Therefore, mapping $x\mapsto\sqrt x$ on the right, we get
$$
\limsup_{x\to\infty}\,xf(x)
=\limsup_{x\to\infty}\tfrac12\left(\cos(x)-\cos(x+2\sqrt x+1)\right)\tag{3 sup}
$$
and
$$
\liminf_{x\to\infty}\,xf(x)
=\liminf_{x\to\infty}\tfrac12\left(\cos(x)-\cos(x+2\sqrt x+1)\right)\tag{3 inf}
$$
When $x$ is big, $\sqrt x$ varies much slower than $x$:
$$
\begin{align}
\sqrt{x+2\pi}-\sqrt x
&=\frac{2\pi}{\sqrt{x+2\pi}+\sqrt x}\\
&\le\frac\pi{\sqrt x}\tag{4}
\end{align}
$$
Thus, for any $\epsilon>0$, choose an $x_0>\dfrac{\pi^2}{\epsilon^2}$ so that $2\sqrt{x_0}+1$ is an odd multiple of $\pi$. Then $(4)$ guarantees that for $x\in[x_0,x_0+2\pi]$, $2\sqrt x+1$ is within $2\epsilon$ of an odd multiple of $\pi$.
Choose the $x\in[x_0,x_0+2\pi]$ which is $0\bmod{2\pi}$ and we get that
$$
\tfrac12\left(\cos(x)-\cos(x+2\sqrt x+1)\right)\ge1-\epsilon\tag{5 sup}
$$
Choose the $x\in[x_0,x_0+2\pi]$ which is $\pi\bmod{2\pi}$ and we get that
$$
\tfrac12\left(\cos(x)-\cos(x+2\sqrt x+1)\right)\le-1+\epsilon\tag{5 inf}
$$
These last two inequalities show that
$$
\limsup_{x\to\infty}\,xf(x)=1\tag{6 sup}
$$
and
$$
\liminf_{x\to\infty}\,xf(x)=-1\tag{6 inf}
$$

A Simpler Approach
As shown in $(2)$,
$$
\begin{align}
xf(x)
&=\frac12\left(\cos\left(x^2\right)-\cos\left((x+1)^2\right)\right)+O\!\left(\frac1x\right)\\
&=\underbrace{\sin\left(x^2+x+\tfrac12\right)}_{\text{frequency $\frac{x+1/2}\pi$}}\underbrace{\sin\left(x+\tfrac12\right)}_{\text{period $2\pi$}}+O\!\left(\frac1x\right)\tag7
\end{align}
$$
So, as $x\to\infty$, the curve looks like a rapidly oscillating sine wave which has an oscillating amplitude with period $2\pi$.

