Prove $\lim_{n \rightarrow \infty} n \int_{\frac{1}{n}}^{1} \frac{\cos(x+\frac{1}{n})-\cos(x)}{x^{\frac{3}{2}}}dx$ exists. Prove $\displaystyle\lim_{n \rightarrow \infty} n \int_{\frac{1}{n}}^{1} \frac{\cos(x+\frac{1}{n})-\cos(x)}{x^{\frac{3}{2}}}\,dx$ exists.
I want to use Dominated convergence theorem to show the limit, but I'm stuck on finding the bound $g(x)$ here. Once I get $g(x)$, I will just exchange the limit and integral.
 A: Applying Cauchy's Mean Value Theorem twice says that there are $0\lt h_1,h_2\lt h$ so that
$$
\begin{align}
\left|\frac1h\left(\frac{\cos(x+h)-\cos(x)}h+\sin(x)\right)\right|
&=\left|\frac{\cos(h)-1}{h^2}\cos(x)+\frac{h-\sin(h)}{h^2}\sin(x)\right|\\
&=\left|-\frac{\cos(h_1)}2\,\cos(x)+\frac{\sin(h_2)}2\sin(x)\right|\\[4pt]
&\le1
\end{align}
$$
For $x\in[h,1]$,
$$
\begin{align}
\left|\frac{\cos(x+h)-\cos(x)}h+\sin(x)\right|\frac1{x^{3/2}}
&\le\left|\frac1h\left(\frac{\cos(x+h)-\cos(x)}h+\sin(x)\right)\right|\frac1{x^{1/2}}\\
&\le\frac1{x^{1/2}}
\end{align}
$$
Therefore,
$$
{\large\chi}_{[h,1]}(x)\left(\frac{\cos(x+h)-\cos(x)}{hx^{3/2}}+\frac{\sin(x)}{x^{3/2}}\right)
$$
is dominated by $\frac1{x^{1/2}}\in L^1[0,1]$. Thus, by Dominated Convergence, we have, with $h=\frac1n$,
$$
\begin{align}
\lim_{n\to\infty}n\int_{1/n}^1\frac{\cos\left(x+\frac1n\right)-\cos(x)}{x^{3/2}}\,\mathrm{d}x
&=\lim_{h\to0}\int_h^1\frac{\cos(x+h)-\cos(x)}{hx^{3/2}}\,\mathrm{d}x\\
&=\lim_{h\to0}\int_0^1{\large\chi}_{[h,1]}(x)\frac{\cos(x+h)-\cos(x)}{hx^{3/2}}\,\mathrm{d}x\\
&=\int_0^1\lim_{h\to0}\,{\large\chi}_{[h,1]}(x)\frac{\cos(x+h)-\cos(x)}{hx^{3/2}}\,\mathrm{d}x\\
&=-\int_0^1\frac{\sin(x)}{x^{3/2}}\,\mathrm{d}x
\end{align}
$$
A: Hint:
Show that 
$$
\int_0^1 \frac{\sin x}{x^{3/2}}dx \quad\text{exists }
$$
And
$$
\lim_{n \rightarrow \infty} n \int_{\frac{1}{n}}^{1} \frac{\cos(x+\frac{1}{n})-\cos(x)}{x^{\frac{3}{2}}}dx=-\int_0^1 \frac{\sin x}{x^{3/2}}dx
$$
