Lebesgue Dominated Convergence Theorem for integral of $\sin$ Let $f$ be continuous strictly positive function with $f(x) \rightarrow 1$
a) For a finite number a >0 find 
$$
\lim_{n\rightarrow \infty} \int_0^a \frac{\sin(x/n)}{x}f(x)dx
$$
b) Find 
$$
\lim_{n\rightarrow \infty} \int_0^\infty \frac{\sin^2(x/n)}{x^2}f(x)dx
$$

for a) it is easy to see that inside the integral it converges to 0 but the only dominating function I find is $|sin(x/n)| \leq x/n $ and as $n\rightarrow \infty$ we have integral is 0? 
not sure if I can do the same for b)
 A: For $a).,\ $ choose $N$ so large that $\frac{a}{N}<\frac{\pi}{2}$ and let $0<M<\infty$ be the bound for $f$ on $\mathbb R^+$, which exists and is finite (why?) and note that $h_n(x):=g_{(N+n)}(x)=\frac{\sin(x/(n+N))}{x}\le \frac{1}{n+N}$ so $h_n$ converges uniformly on $[0,a]$. Hence, 
$\lim_{n\rightarrow \infty} \int_0^a \frac{\sin(x/n)}{x}f(x)dx\le M\lim_{n\rightarrow \infty} \int_0^a h_n(x)dx=M \int_0^a \lim_{n\rightarrow \infty}h_n(x)dx=0.$
For $b).,\ $ write $\int_0^\infty \frac{\sin^2(x/n)}{x^2}f(x)dx=\int_0^1 \frac{\sin^2(x/n)}{x^2}f(x)dx+\int_1^\infty \frac{\sin^2(x/n)}{x^2}f(x)dx$.
Now $\frac{\sin^2(x/n)}{x^2}=\frac{\sin^2(x/n)}{n^2(x/n)^2}$ so the integrand in the first integral is dominated by $h(x)=2M$ and the integrand in the second integral is dominated by $h(x)=\frac{M}{x^2}$ and so an application of the DCT shows that each of these integrals is $0.$
A: First of all, $ \left(\forall x\in\mathbb{R}\right),\ \left|\frac{\sin{x}}{x}\right|=\left|\int_{0}^{1}{\cos{\left(xy\right)}\,\mathrm{d}y}\right|\leq \int_{0}^{1}{\left|\cos{\left(xy\right)}\right|\mathrm{d}y}\leq 1 \cdot $
And $ \left(\forall x\in\mathbb{R}\right),\ \left|\frac{\sin^{2}{x}}{x^{2}}\right|=\left|\int_{0}^{1}{\frac{\sin{\left(2xy\right)}}{x}\,\mathrm{d}y}\right|\leq\int_{0}^{1}{\left|\frac{\sin{\left(2xy\right)}}{x}\right|\mathrm{d}y}\leq \int_{0}^{1}{2y\,\mathrm{d}y}=1 \cdot $
Thus \begin{aligned}\left|\int_{0}^{a}{\frac{\sin{\left(\frac{x}{n}\right)}}{x}f\left(x\right)\mathrm{d}x}\right|\leq\int_{0}^{a}{\left|\frac{\sin{\left(\frac{x}{n}\right)}}{x}f\left(x\right)\right|\mathrm{d}x}\leq\frac{1}{n}\int_{0}^{a}{\left|f\left(x\right)\right|\mathrm{d}x}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}
And \begin{aligned}\left|\int_{0}^{a}{\frac{\sin^{2}{\left(\frac{x}{n}\right)}}{x^{2}}f\left(x\right)\mathrm{d}x}\right|\leq\int_{0}^{a}{\left|\frac{\sin^{2}{\left(\frac{x}{n}\right)}}{x^{2}}f\left(x\right)\right|\mathrm{d}x}\leq\frac{1}{n^{2}}\int_{0}^{a}{\left|f\left(x\right)\right|\mathrm{d}x}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}
