<s>[INST] 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$, $xf(x)=±\sin(2n\pi+\frac{\pi}{2})=±1$.
 [/INST] Here is a clarified version of your proof. (In fact, it was a part of my original solution, but I modified it by following your idea.)
Step 1. Estimation of $xf(x)$

Let $t = \sqrt{u}$. Then
\begin{align*}
xf(x)
&= x\int_{x}^{x+1} \sin (t^2) \, dt = x \int_{x^2}^{(x+1)^2}  \frac{\sin u}{2\sqrt{u}} \, du \\
&= x \left[ \frac{1-\cos u}{2\sqrt{u}} \right]_{x^2}^{(x+1)^2} + x \int_{x^2}^{(x+1)^2}  \frac{1-\cos u}{4u^{3/2}} \, du
\end{align*}
Since
\begin{align*}
x \left[ \frac{1-\cos u}{2\sqrt{u}} \right]_{x^2}^{(x+1)^2}
&= \frac{x}{2} \left(\frac{1-\cos\big((x+1)^2\big)}{x+1} - \frac{1 - \cos\big(x^2\big)}{x} \right) \\
&= \frac{1}{2} \left(\cos \left( x^2 \right) - \cos\left((x+1)^2\right) \right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)} \\
&= \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)},
\end{align*}
we have 
$$xf(x) = \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)} + x \int_{x^2}^{(x+1)^2}  \frac{1-\cos u}{4u^{3/2}} \, du. $$
It is easy to observe that 
$$ - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)} + x \int_{x^2}^{(x+1)^2}  \frac{1-\cos u}{4u^{3/2}} \, du = O\left(\frac{1}{x}\right). $$
Indeed, it follows from the estimation
\begin{align*}
\left| x \int_{x^2}^{(x+1)^2}  \frac{1-\cos u}{4u^{3/2}} \, du \right|
&\leq x \int_{x^2}^{(x+1)^2}  \frac{\left| 1-\cos u \right| }{4u^{3/2}} \, du \\
&\leq x \int_{x^2}^{(x+1)^2}  \frac{1}{2x^3} \, du
 \leq \frac{2x+1}{2x^2}
 = O\left(\frac{1}{x}\right).
\end{align*}
Step 2. Evaluation of limit superior and limit inferior

The estimation above in particular implies that
$$ \limsup_{x\to\infty} xf(x) = \limsup_{x\to\infty} \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right)$$ 
and likewise for the liminf. To find the limsup, note from the identity above that 
$$\limsup_{x\to\infty} xf(x) \leq 1. $$
We claim that it is indeed 1. To this end, it suffices to find a subsequence $(x_n)$ such that $x_n \to \infty$ and
$$ \sin \left(x_n +\frac{1}{2}\right)\sin\left(x_n^2+x_n+\frac{1}{2}\right) \to 1. $$
Let $x = \sqrt{2\pi k}$. Then we have
$$ \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right) = \sin^2 \left(\sqrt{2\pi k}+\frac{1}{2}\right) $$
As $\sqrt{2\pi k} \to \infty$ and $\sqrt{2\pi(k+1)} - \sqrt{2\pi k} \to 0$ as $k \to \infty$, we can find a subsequence $(k_n)$ such that
$$ \min \left\{ \left| \sqrt{2\pi k_n} + \frac{1}{2} - \left( m+\frac{1}{2}\right)\pi \right| : m \in \Bbb{Z} \right\} \to 0 \quad \text{as} \quad n \to \infty. \tag{1}$$
Thus for $x_n = \sqrt{2\pi k_n}$ we obtain $ x_n f(x_n) \to 1$ as $n \to \infty$ and therefore $\limsup_{x\to\infty} xf(x) = 1$. A slight modification of this argument also proves that $\liminf_{x\to\infty} xf(x) = -1$.
Proof of the claim $(1)$

Let $\epsilon > 0$. Then there exists $N$ such that whenever $k \geq N$ we have $0 < r_k < \epsilon$, where $r_k$ denotes $r_k = \sqrt{2\pi(k+1)} - \sqrt{2\pi k}$ for simplicity.  Then for the sequence of open balls 
$$ B_k = \left(\sqrt{2\pi k} - r_k, \sqrt{2\pi k} + r_k\right) = \left(\sqrt{2\pi k} - r_k, \sqrt{2\pi (k+1)}\right), $$
we have $B_k \cap B_{k+1} \neq \varnothing $ and hence
$$ \bigcup_{k=N}^{\infty} B_k = \left(\sqrt{2\pi N} - r_{N}, \infty \right). $$
Here, we used the fact that $\sqrt{2\pi k} \to \infty$ as $k \to \infty$. Then for any sufficiently large integer $m$ we have
$$ \left(m+\frac{1}{2}\right)\pi - \frac{1}{2} \in \left(\sqrt{2\pi N} - r_{N}, \infty \right). $$
Thus we can pick some $k \geq N$ satisfing
$$ \left(m+\frac{1}{2}\right)\pi - \frac{1}{2} \in B_k \subset \left(\sqrt{2\pi N} - \epsilon, \sqrt{2\pi N} + \epsilon \right). $$
This proves the following proposition: For any $\epsilon > 0$ there exists a positive integer $k$ such that
$$ \min \left\{ \left| \sqrt{2\pi k} + \frac{1}{2} - \left( m+\frac{1}{2}\right)\pi \right| : m \in \Bbb{Z} \right\} < \epsilon.$$
Then the claim $(1)$ immediately follows.
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